The `v–S` graph for a test vehicle is shown.

Determine its acceleration when `S = 100\ m` and when `S = 175\ m`.

Slope of the velocity line

\begin{aligned} slope &= \frac{\Delta v}{\Delta S} \\[1pt] slope &= \frac{0 - 50}{200 - 150} \\[1pt] slope &= -1 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] v_2 &= (slope)S + b \\[1pt] 0 &= (-1)(200) + b \\[1pt] b &= 200 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_2 &= (slope)S + b \\[1pt] v_2 &= (-1)S + 200 \\[1pt] v_2 &= -S + 200 \\[1pt] \end{aligned}

The `v–S` graph of a cyclist traveling along a straight road is shown.

Construct the `a–S` graph.

The boat travels along a straight line with the speed described by the graph.

Construct the `S–t` and `a–S` graphs.

Also, determine the time required for the boat to travel a distance `S = 400\ m` if `S = 0` when `t = 0`.

\begin{aligned} t &= t_1(100) + t_2(400) \\[1pt] t &= \sqrt{100} + \frac{1}{0.2} \left(\ln \left(400\right) - \ln \left(100\right)\right) \\[1pt] t &= 10 + 6.93 \\[1pt] t &= 16.93\ s \\[1pt] \end{aligned}

Time required for the boat to travel a distance `400\ m` is `\boxed{16.93\ seconds}`

The jet plane starts from rest at `S = 0` and is subjected to the acceleration shown.

Determine the speed of the plane when it has traveled `1000\ ft`.

Also, how much time is required for it to travel `1000\ ft`?

`v(1000)` is the speed of the plane when it has traveled `1000\ ft`.

\begin{aligned} v(1000) &= \sqrt{150(1000) - 0.025(1000)^2} \\[1pt] v(1000) &= 353.55\ ft/S \\[1pt] \end{aligned}

\begin{aligned} v &= \sqrt{150S - 0.025S^2} \\[1pt] \frac{dS}{dt} &= \sqrt{150S - 0.025S^2} \\[1pt] \int_0^t dt &= \int_0^{S} \frac{dS}{\sqrt{150S - 0.025S^2}} \\[1pt] t &= \int_0^{S} \frac{dS}{\sqrt{150S - 0.025S^2}} \\[1pt] t &= \frac{1}{\sqrt{0.025}} \int_0^{S} \frac{dS}{\sqrt{6000S - S^2}} \\[1pt] \end{aligned}

Simplify denominator (completing the square)

\begin{aligned} -S^2 + 6000S \\[1pt] - \left(S^2 - 6000S\right) \\[1pt] - \left(S^2 + 2(-3000)S + (-3000)^2 - (-3000)^2\right) \\[1pt] - \left(\left(S - 3000\right)^2 - (-3000)^2\right) \\[1pt] (-3000)^2 - \left(S - 3000\right)^2 \\[1pt] (3000)^2 - \left(S - 3000\right)^2 \\[1pt] \end{aligned}

Substitute simplied denominator

\begin{aligned} t &= \frac{1}{\sqrt{0.025}} \int_0^{1000} \frac{dS}{\sqrt{(3000)^2 - \left(S - 3000\right)^2}} \\[1pt] \end{aligned}

`u`-substitution

\begin{aligned} u &= S - 3000\\[1pt] \frac{du}{dS} &= 1\\[1pt] du &= dS\\[1pt] \end{aligned}

Apply `u`-substitution and solve the integral

\begin{aligned} t &= \frac{1}{\sqrt{0.025}} \int_0^{1000} \frac{du}{\sqrt{(3000)^2 - \left(u\right)^2}} \\[1pt] t &= \frac{1}{\sqrt{0.025}} \left[\arcsin \left(\frac{u}{3000}\right)\right]_0^{1000} \\[1pt] t &= \frac{1}{\sqrt{0.025}} \left[\arcsin \left(\frac{S - 3000}{3000}\right)\right]_0^{1000} \color{green}\text{ //substituted }u \\[1pt] t &= -4.62 - \left(-9.93\right) \\[1pt] t &= 5.32\ s \\[1pt] \end{aligned}

The motion of a train is described by the `a–S` graph shown.

Draw the `v–S` graph if `v = 0` at `S = 0`.

Slope of acceleration line

\begin{aligned} jerk &= \frac{\Delta a}{\Delta S} \\[1pt] jerk &= \frac{3 - 0}{300 - 0} \\[1pt] jerk &= \frac{3}{300} \\[1pt] jerk &= 0.01 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_1 &= (jerk)S \\[1pt] a_1 &= 0.01S \\[1pt] \end{aligned}

Slope of acceleration line

\begin{aligned} jerk &= \frac{\Delta a}{\Delta S} \\[1pt] jerk &= \frac{0 - 3}{600 - 300} \\[1pt] jerk &= \frac{-3}{300} \\[1pt] jerk &= -0.01 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] a_2 &= (jerk)(S) + b \\[1pt] 0 &= (-0.01)(600) + b \\[1pt] b &= 6 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_2 &= (jerk)(S) + b \\[1pt] a_2 &= -0.01S + 6 \\[1pt] \end{aligned}

From experimental data, the motion of a jet plane while traveling along a runway is defined by the `v–t` graph.

Construct the `S–t` and `a–t` graphs for the motion.

When `t = 0, S = 0`.

Two cars start from rest side by side and travel along a straight road.

Car `A` accelerates at `4\ m/s^2` for `10 s` and then maintains a constant speed.

Car `B` accelerates at `5\ m/s^2` until reaching a constant speed of `25\ m/s` and then maintains this speed.

Construct the `a–t, v–t` and `S–t` graphs for each car until `t = 15 s`.

What is the distance between the two cars when `t = 15 s`?

It's given that after `10\ seconds`, Car `A` maintains a constant speed.

Following is the velocity function of Car `A`, after it stops accelerating.

\begin{aligned} v &= u + at \\[1pt] v_{A_2} &= v_{A_1}(10) + (0)t \\[1pt] v_{A_2} &= 40 \\[1pt] \end{aligned}

Following is the position function of Car `A`, after it stops accelerating.

\begin{aligned} v_{A_2} &= 40 \\[1pt] \frac{dS}{dt} &= 40 \\[1pt] \int_{S_{A_1}(10)}^{S_{A_2}} dS &= \int_{10}^t 40\ dt \\[1pt] S_{A_2} - S_{A_1}(10) &= 40 t - 40(10) \\[1pt] S_{A_2} &= 40 t - 400 + 200 \\[1pt] S_{A_2} &= 40 t - 200 \\[1pt] \end{aligned}

It's given that Car `B` accelerates until reaching a constant speed of `25\ m/s` and then maintains this speed.

Therefore, the function that describe the velocity of Car `B` during this interval is following:

\begin{aligned} v_{B_2} &= 25 \\[1pt] \end{aligned}

Now, derive the position function for Car `B` after it stops accelerating is i.e. which it reaches the constant speed of `25\ m/s`.

How long does it take for Car `B` to reach this constant speed?

\begin{aligned} v_{B_1} &= 5t \\[1pt] 25 &= 5t \\[1pt] t &= 5 \\[1pt] \end{aligned}

Car `B` takes `\boxed{5\ seconds}` to reach the constant speed of `25\ m/s`.

Position function

The position function of Car `B` after it stops accelerating is following:

\begin{aligned} v_{B_2} &= 25 \\[1pt] \frac{dS}{dt} &= 25 \\[1pt] \int_{S_{B_1}(5)}^{S_{B_2}} dS &= \int_5^t 25\ dt \\[1pt] S_{B_2} - S_{B_1}(5) &= \Big[25t\Big]_5^t \\[1pt] S_{B_2} - 62.5 &= 25t - 125 \\[1pt] S_{B_2} &= 25t - 125 + 62.5 \\[1pt] S_{B_2} &= 25t -62.5 \\[1pt] \end{aligned}

`S_{distance}` is the distance between the two cars when `t = 15 s`.

\begin{aligned} S_{distance} &= S_{A_2}(15) - S_{B_2}(15) \\[1pt] S_{distance} &= 400 - 312.5 \\[1pt] S_{distance} &= 87.5\ m \\[1pt] \end{aligned}

A man riding upward in a freight elevator accidentally drops a package off the elevator when it is `100\ ft` from the ground.

If the elevator maintains a constant upward speed of `4\ ft/s`, determine how high the elevator is from the ground the instant the package hits the ground.

Draw the `v–t` curve for the package during the time it is in motion.

Assume that the package was released with the same upward speed as the elevator.

Convert gravity into `ft/s^2`

\begin{aligned} g &= -9.81 \ m/s^2 \\[1pt] g &= \frac{-9.81}{0.3048}\ ft/s^2 = \boxed{-32.19 \ ft/s^2} \\[1pt] \end{aligned}

What is the final velocity of the package when it hits the ground?

\begin{aligned} v^2 &= u^2 + 2aS \\[1pt] v^2 &= 4^2 + 2(-32.19)(0 - 100) \\[1pt] v &= 80.33\ ft/s \\[1pt] \end{aligned}

The package hits the ground with velocity `\boxed{80.33\ ft/s}`.

How long does the package take to reach its final velocity?

\begin{aligned} v &= u + at \\[1pt] -80.33 &= 4 - 9.81t \\[1pt] t &= 2.62 \\[1pt] \end{aligned}

It takes `\boxed{2.62\ seconds}` for the package to hit the ground.

How far did the elevator travel up during those `2.62\ seconds`?

It's given that the elevator maintains a constant upward speed of `4\ ft/s`.

\begin{aligned} S_{fall} &= 4 \times 2.62 \\[1pt] S_{fall} &= 10.48 \\[1pt] \end{aligned}

The elevator travelled up a distance of `\boxed{10.48\ ft}` while the package was falling.

What is the total distance travelled by the elevator?

\begin{aligned} S_{distance} &= 100 + S_{fall} \\[1pt] S_{distance} &= 100 + 10.48 \\[1pt] S_{distance} &= 110.48\ ft \\[1pt] \end{aligned}

The elevator travelled up a total distance of `\boxed{110.48\ ft}` from the ground.

The speed of a train during the first minute has been recorded as follows:

Plot the `v–t` graph, approximating the curve as straight-line segments between the given points.

Determine the total distance traveled.

Distance travelled in first interval `t \in [0, 20]` is area of the triangle.

\begin{aligned} S_1 &= \frac{1}{2} (20 - 0) (16) \\[1pt] S_1 &= 160\ m \\[1pt] \end{aligned}

Distance travelled in second interval `t \in [20, 40]` is area of the trapezoid.

\begin{aligned} S_2 &= \frac{1}{2} (16 + 21) (40 - 20) \\[1pt] S_2 &= 370\ m \\[1pt] \end{aligned}

Distance travelled in third interval `t \in [40, 60]` is area of the trapezoid.

\begin{aligned} S_3 &= \frac{1}{2} (21 + 24) (60 - 40) \\[1pt] S_3 &= 450\ m \\[1pt] \end{aligned}

A two-stage rocket is fired vertically from rest with the acceleration shown.

After `15 s` the first stage `A` burns out and the second stage `B` ignites.

Plot the `v–t` and `S–t` graphs which describe the motion of the second stage for `0 \le t \le 40s`.

Acceleration here is directly proportional to `t`.

\begin{aligned} a_1 &= t \\[1pt] \end{aligned}

Velocity function can be found using either one of the following approaches.

Using integration

\begin{aligned} a_2 &= 20 \\[1pt] \frac{dv}{dt} &= 20 \\[1pt] \int_{v_1(15)}^{v_2} dv &= \int_{15}^t 20\ dt \\[1pt] v_2 - v_1(15) &= 20t - 20(15) \\[1pt] v_2 &= 20t - 300 + 112.5 \\[1pt] v_2 &= 20t -187.5 \\[1pt] \end{aligned}

Using equation of motion with constant acceleration

\begin{aligned} v &= u + at \\[1pt] v_2 &= v_1(15) + (20)(t - 15) \\[1pt] v_2 &= 112.5 + 20t - 20(15) \\[1pt] v_2 &= 112.5 + 20t - 300 \\[1pt] v_2 &= 20t -187.5 \\[1pt] \end{aligned}

Starting from rest at `S = 0`, a boat travels in a straight line with the acceleration shown by the `a–S` graph.

12-56

Determine the boat’s speed when `S = 50\ ft`, `100\ ft`, and `150\ ft`.

12-57

Construct the v–s graph.

Slope of the acceleration line

\begin{aligned} jerk &= \frac{\Delta a}{\Delta S} \\[1pt] jerk &= \frac{6 - 8}{100 - 0} \\[1pt] jerk &= -0.02 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] 8 &= (-0.02)(0) + b \\[1pt] b &= 8 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} y &= mx + b \\[1pt] a_1 &= (jerk)S + b \\[1pt] a_1 &= -0.02 S + 8 \\[1pt] \end{aligned}

\begin{aligned} a_1 &= -0.02 S + 8 \\[1pt] v \frac{dv}{dS} &= -0.02 S + 8 \\[1pt] \int_0^{v_1} v \ dv &= \int_0^S \left(-0.02 S + 8\right)\ dS \\[1pt] \frac{\left(v_1\right)^2}{2} &= \left[-0.01 S^2 + 8S\right]_0^S \\[1pt] \frac{\left(v_1\right)^2}{2} &= -0.01 S^2 + 8S \\[1pt] v_1 &= \sqrt{-0.02 S^2 + 16S} \\[1pt] \end{aligned}

To determine the boat’s speed when `S = 50\ ft` and `S = 100\ ft` execute the function `v_1` for those `S` values i.e. `v_1(50)` and `v_1(100)`.

Slope of the acceleration line

\begin{aligned} jerk &= \frac{\Delta a}{\Delta S} \\[1pt] jerk &= \frac{0 - 6}{150 - 100} \\[1pt] jerk &= -0.12 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] a &= (jerk)S + b \\[1pt] 0 &= (-0.12)(150) + b \\[1pt] b &= 18 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} y &= mx + b \\[1pt] a_2 &= (jerk)S + b \\[1pt] a_2 &= -0.12 S + 18 \\[1pt] \end{aligned}

\begin{aligned} a_2 &= -0.12 S + 18 \\[1pt] v \frac{dv}{dS} &= -0.12 S + 18 \\[1pt] \int_{v_1(100)}^{v_2} v \ dv &= \int_{100}^S \left(-0.12 S + 18\right)\ dS \\[1pt] \left[\frac{v^2}{2}\right]_{v_1(100)}^{v_2} &= \left[-0.06 S^2 + 18 S\right]_{100}^S \\[1pt] \frac{\left(v_2\right)^2}{2} - \frac{\left(v_1(100)\right)^2}{2} &= -0.06 S^2 + 18 S + 0.06(100)^2 - 18(100) \\[1pt] \left(v_2\right)^2 &= -0.12 S^2 + 36 S + 0.12(100)^2 - 36(100) + \left(37.42\right)^2 \\[1pt] v_2 &= \sqrt{-0.12 S^2 + 36 S -1000} \\[1pt] \end{aligned}

To determine the boat’s speed when `S = 150\ ft` execute the function `v_2` for that `S` value i.e. `v_2(150)`.

An airplane lands on the straight runway, originally traveling at `110\ ft/s` when `S = 0`.

If it is subjected to the decelerations shown, determine the time `t'` needed to stop the plane and construct the `S–t` graph for the motion.

`t'` is when `v_4 = 0`

\begin{aligned} v_4 &= -3t + 100 \\[1pt] -3t' + 100 &= 0 \\[1pt] t' &= \frac{100}{3} \\[1pt] t' &= 33.33\ s \\[1pt] \end{aligned}

The `v–t` graph for the motion of a car as it moves along a straight road is shown.

Draw the `S–t` and `a–t` graphs.

Also determine the average speed and the distance traveled for the `15\ s` time interval.

When `t = 0`, `S = 0`.

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] v_2 &= a_2t + b \color{green}\text{ //function 1}\\[1pt] 0 &= (-1.5)(15) + b \\[1pt] b &= 22.5 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_2 &= a_2t + b \color{green}\text{ //function 1}\\[1pt] v_2 &= -1.5t + 22.5 \\[1pt] \end{aligned}

`S_{distance}` is the distance traveled for the `15\ s` time interval.

\begin{aligned} S_{distance} &= \left|\int_0^5 v_1\ dt\right| + \left|\int_{5}^{15} v_2\ dt\right| \\[1pt] S_{distance} &= \left|S_1(5) - S_1(0)\right| + \left|S_2(15) - S_2(5)\right| \\[1pt] S_{distance} &= \left|25 - 0\right| + \left|100 - 25\right| \\[1pt] S_{distance} &= 100\ m \\[1pt] \end{aligned}

A motorcycle starts from rest at `S = 0` and travels along a straight road with the speed shown by the `v–t` graph.

12-52

• Determine the total distance the motorcycle travels until it stops when `t = 15\ s`.
• Also plot the `a–t` and `S–t` graphs.
  

12-53

Determine the motorcycle’s acceleration and position when `t = 8\ s` and `t = 12\ s`.

`S_{total}` is the total distance the motorcycle travels until it stops at `t = 15\ s`.

\begin{aligned} S_{total} &= \left|\int_0^4 v_1\ dt\right| + \left|\int_{4}^{10} v_2\ dt\right| + \left|\int_{10}^{15} v_3\ dt\right| \\[1pt] S_{total} &= \left|S_1(4) - S_1(0)\right| + \left|S_2(10) - S_2(4)\right| + \left|S_3(15) - S_2(10)\right| \\[1pt] S_{total} &= \left|10 - 0\right| + \left|40 - 10\right| + \left|52.5 - 40\right| \\[1pt] S_{total} &= 52.5\ m \\[1pt] \end{aligned}

The `v–t` graph for a train has been experimentally determined.

From the data, construct the `S–t` and `a–t` graphs for the motion for `0 \le t \le 180\ s`.

When `t = 0, S = 0`.

Velocity function for the first interval.

\begin{aligned} v_1 &= \frac{\Delta v}{\Delta t}t \\[1pt] v_1 &= \left(\frac{6 - 0}{60 - 0}\right)t \\[1pt] v_1 &= 0.1t \\[1pt] \end{aligned}

Position function for the first interval.

\begin{aligned} S_1 &= \int_0^t 0.1t \\[1pt] S_1 &= 0.05t^2 \\[1pt] \end{aligned}

Acceleration function for the first interval.

\begin{aligned} a_1 &= \frac{d}{dt}\left[v_1\right] \\[1pt] a_1 &= \frac{d}{dt}\left[0.1t\right] \\[1pt] a_1 &= 0.1 \\[1pt] \end{aligned}

Velocity function for the second interval.

\begin{aligned} v_2 &= 6 \\[1pt] \end{aligned}

Position function for the second interval.

\begin{aligned} v_2 &= 6 \\[1pt] \frac{dS}{dt} &= 6 \\[1pt] \int_{S_1(60)}^{S_2} dS &= \int_{60}^t 6\ dt \\[1pt] S_2 - S_1(60) &= \Big[6t\Big]_{60}^t \\[1pt] S_2 - 180 &= 6t - 360 \\[1pt] S_2 &= 6t -180 \\[1pt] \end{aligned}

Slope of the velocity curve/incline

\begin{aligned} slope &= \frac{\Delta v}{\Delta t} \\[1pt] slope &= \frac{10 - 6}{180 - 120} \\[1pt] slope &= 0.07 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] 6 &= (0.07)(120) + b \\[1pt] b &= -2 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_3 &= (slope)t + b \\[1pt] v_3 &= 0.07t -2 \\[1pt] \end{aligned}

The car starts from rest at `S = 0` and is subjected to an acceleration shown by the `a–S` graph.

Draw the `v–S` graph and determine the time needed to travel `200\ ft`.

Acceleration function for the first interval.

\begin{aligned} a_1 &= 12 \\[1pt] \end{aligned}

Velocity function for the first interval.

\begin{aligned} a_1 &= 12 \\[1pt] v \frac{dv}{dS} &= 12 \\[1pt] \int_0^{v_1} v\ dv &= \int_0^S 12\ dS \\[1pt] \left[\frac{v^2}{2}\right]_0^{v_1} &= 12S \\[1pt] \frac{\left(v_1\right)^2}{2} &= 12S \\[1pt] v_1 &= \sqrt{24S} \\[1pt] \end{aligned}

Position function for the first interval.

\begin{aligned} v_1 &= \sqrt{24S} \\[1pt] \frac{dS}{dt} &= \sqrt{24S} \\[1pt] \frac{dS}{\sqrt{24S}} &= dt \\[1pt] \frac{1}{\sqrt{24}}\int_0^{S_1} S^{-1/2}\ dS &= \int_0^t dt \\[1pt] \frac{1}{\sqrt{24}} \left[2\sqrt{S}\right]_0^{S_1} &= t \\[1pt] \frac{2}{\sqrt{24}} \sqrt{S_1} &= t \color{green}\text{ //equation 1} \\[1pt] \sqrt{S_1} &= \frac{t\sqrt{24}}{2} \\[1pt] S_1 &= 8t^2 \\[1pt] \end{aligned}

`t_{200ft}` is the time taken to travel `200\ ft`.

\begin{aligned} t &= \frac{2}{\sqrt{24}} \sqrt{S_1} \color{green}\text{ //equation 1} \\[1pt] t_{200ft} &= \frac{2}{\sqrt{24}} \sqrt{200} \\[1pt] t_{200ft} &= 5.77\ s \\[1pt] \end{aligned}

The jet car is originally traveling at a velocity of `10\ m/s` when it is subjected to the acceleration shown.

Determine the car’s maximum velocity and the time `t'` when it stops.

When `t = 0, S = 0`.

Velocity function for the first interval.

\begin{aligned} v &= u + at \\[1pt] v_1 &= 10 + 6t \\[1pt] \end{aligned}

Velocity function for the second interval.

\begin{aligned} v &= u + at \\[1pt] v_2 &= v_1(15) + (-4)t \\[1pt] v_2 &= 100 - 4t \\[1pt] \end{aligned}

Time velocity is `0` in the second interval

\begin{aligned} v_2 &= 0 \\[1pt] 100 - 4t_2 &= 0 \\[1pt] t_2 &= \frac{100}{4} \\[1pt] t_2 &= 25 \\[1pt] \end{aligned}

`t'` is `t_2` plus the time taken to complete the first interval

\begin{aligned} t' &= t_2 + 15 \\[1pt] t' &= 25 + 15 \\[1pt] t' &= 40 \\[1pt] \end{aligned}

The race car starts from rest and travels along a straight road until it reaches a speed of `26\ m/s` in `8\ s` as shown on the `v–t` graph.

The flat part of the graph is caused by shifting gears.

Draw the `a–t` graph and determine the maximum acceleration of the car.

Acceleration function for the first interval when `t \in [0, 4]`

\begin{aligned} a_1 &= \frac{dv}{dt} \\[1pt] a_1 &= \frac{d}{dt} \left[3.5t\right] \\[1pt] a_1 &= 3.5\ m/s^2 \\[1pt] \end{aligned}

Acceleration function for the second interval when `t \in [4, 5]`

\begin{aligned} a_2 &= \frac{dv}{dt} \\[1pt] a_2 &= \frac{d}{dt} \left[14\right] \\[1pt] a_2 &= 0 \\[1pt] \end{aligned}

Acceleration function for the third interval when `t \in [5, 8]`

\begin{aligned} a_3 &= \frac{dv}{dt} \\[1pt] a_3 &= \frac{d}{dt} \left[4t - 6\right] \\[1pt] a_3 &= 4\ m/s^2 \\[1pt] \end{aligned}

A two-stage rocket is fired vertically from rest at `S = 0` with the acceleration as shown.

After `30\ s` the first stage, `A`, burns out and the second stage, `B`, ignites.

Plot the `v–t` and `S–t` graphs which describe the motion of the second stage for `0 \le t \le 60\ s`.

First, calculate the acceleration function during the first interval.

Slope

\begin{aligned} slope &= \frac{\Delta a}{\Delta t} \\[1pt] slope &= \frac{12 - 0}{30 - 0} \\[1pt] slope &= 0.4 \\[1pt] \end{aligned}

`a` intercept

\begin{aligned} y &= mx + b \\[1pt] 12 &= (0.4)(30) + b \\[1pt] b &= 0 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} y &= mx + b \\[1pt] a_1 &= 0.4t + 0 \\[1pt] a_1 &= 0.4t \\[1pt] \end{aligned}

The `a–S` graph for a rocket moving along a straight track has been experimentally determined.

If the rocket starts at `S = 0` when `v = 0`, determine its speed when it is at `S = 75\ ft`, and `125\ ft`, respectively.

Use Simpson’s rule with `n = 100` to evaluate `v` at `S = 125\ ft`.

Velocity function

\begin{aligned} a &= 5 \\[1pt] v \frac{dv}{dS} &= 5 \\[1pt] \int_0^{v_1} v\ dv &= \int_0^S 5\ dS \\[1pt] \left[\frac{v^2}{2}\right]_0^{v_1} &= \Big[5S\Big]_0^S \\[1pt] \frac{\left(v_1\right)^2}{2} &= 5S \\[1pt] v_1 &= \sqrt{10S} \\[1pt] \end{aligned}

The `v–t` graph for a particle moving through an electric field from one plate to another has the shape shown in the figure.

Where `t' = 0.2\ s` and `v_{max} = 10\ m/s`.

Draw the `S–t` and `a–t` graphs for the particle.

When `\displaystyle{t = \frac{t'}{2}}` the particle is at `S = 0.5\ m`.

It's given that `t' = 0.2`, therefore `t_{mid}` is:

\begin{aligned} t_{mid} &= \frac{t'}{2} \\[1pt] t_{mid} &= \frac{0.2}{2} \\[1pt] t_{mid} &= 0.1 \\[1pt] \end{aligned}

First, calculate the constant acceleration until the mid of the electric field.

\begin{aligned} v &= u + at \\[1pt] v_{max} &= u + \left(a_1\right) \left(t_{mid}\right) \\[1pt] 10 &= 0 + \left(a_1\right)(0.1) \\[1pt] a_1 &= \frac{10}{0.1} \\[1pt] a_1 &= 100 \\[1pt] \end{aligned}

The particle is deceleration now.

\begin{aligned} a_2 &= -100 \\[1pt] \end{aligned}

The `v–t` graph for a particle moving through an electric field from one plate to another has the shape shown in the figure.

The acceleration and deceleration that occur are constant and both have a magnitude of `4\ m/s^2`.

If the plates are spaced `200\ mm` apart, determine the maximum velocity `v_{max}` and the time `t'` for the particle to travel from one plate to the other.

Also draw the `S–t` graph.

When `\displaystyle{t = \frac{t'}{2}}` the particle is at `S = 100\ mm`.

Acceleration function

\begin{aligned} a_1 &= 4 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_1 &= \int_0^t a_1 \ dt \\[1pt] v_1 &= 4t \\[1pt] \end{aligned}

Displacement function

\begin{aligned} S_1 &= \int_0^t 4t \ dt \\[1pt] S_1 &= 2t^2 \\[1pt] \end{aligned}

Time `\left(t_{mid}\right)` taken to reach the middle of the electric field i.e. at `\displaystyle{\frac{200\ mm}{2} = \frac{0.2\ meters}{2}} = 0.1\ meters` is:

\begin{aligned} S_1 &= 2t^2 \\[1pt] 0.1 &= 2 \left(t_{mid}\right)^2 \\[1pt] t_{mid} &= \sqrt{\frac{0.1}{2}} \\[1pt] t_{mid} &= 0.22\ s \\[1pt] \end{aligned}

Since the magnitudes of acceleration and deceleration are the same, time `\left(t'\right)` to reach the end plate is two times the time `\left(t_{mid}\right)` taken to reach the electric field:

\begin{aligned} t' &= 2 \times t_{mid} \\[1pt] t' &= 2 \times 0.22 \\[1pt] t' &= 0.45\ s \\[1pt] \end{aligned}

`v_{max}` is the velocity `v_1` of the particle at the middle of the electric field i.e. when time is `t_{mid}`.

\begin{aligned} v_1 &= 4t \\[1pt] v_{max} &= 4 \left(t_{mid}\right) \\[1pt] v_{max} &= 4 \left(0.22\right) \\[1pt] v_{max} &= 0.89\ m/s \\[1pt] \end{aligned}

`S_{mid}` is the distance to middle of the electric field.

\begin{aligned} S_{mid} &= S_1(t_{mid}) \\[1pt] S_{mid} &= S_1(0.22) \\[1pt] S_{mid} &= 0.1\ m \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_2 &= -4 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} a_2 &= -4 \\[1pt] \frac{dv}{dt} &= -4 \\[1pt] \int_{v_{max}}^{v_2} dv &= \int_{t_{mid}}^{t} -4\ dt \\[1pt] v_2 - v_{max} &= \Big[-4t\Big]_{t_{mid}}^{t} \\[1pt] v_2 - v_{max} &= -4t + 4\left(t_{mid}\right) \\[1pt] v_2 - 0.89 &= -4t + 4\left(0.22\right) \\[1pt] v_2 &= -4t + 1.79 \\[1pt] \end{aligned}

Displacement function

\begin{aligned} v_2 &= -4t + 1.79 \\[1pt] \frac{dS}{dt} &= -4t + 1.79 \\[1pt] \int_{S_{mid}}^{S_2} dS &= \int_{t_{mid}}^{t} \left(-4t + 1.79\right)\ dt \\[1pt] S_2 - S_{mid} &= \left[-2t^2 + 1.79t\right]_{t_{mid}}^{t} \\[1pt] S_2 &= -2t^2 + 1.79t + 2 \left(0.22\right)^2 - 1.79(0.22) + 0.1 \\[1pt] S_2 &= -2t^2 + 1.79t -0.2 \\[1pt] \end{aligned}

The motion of a jet plane just after landing on a runway is described by the `a–t` graph.

Determine the time `t'` when the jet plane stops.

Construct the `v–t` and `S–t` graphs for the motion.

Here `S = 0`, and `v = 300\ ft/s` when `t = 0`.

For the first `10\ seconds` the acceleration is `0`.

\begin{aligned} a_1 &= 0 \\[1pt] \end{aligned}

Initial velocity at `t = 0` is given to be `300\ ft/s` which is `91.44\ m/s`.

Since there is no acceleration, velocity is constant throughout this interval.

\begin{aligned} v_1 &= 91.44 \\[1pt] \end{aligned}

Slope of acceleration graph during the interval `t \in [10, 20]`

\begin{aligned} slope &= \frac{\Delta a}{\Delta t} \\[1pt] slope &= \frac{-10 - (-20)}{20 - 10} \\[1pt] slope &= \frac{10}{10} \\[1pt] slope &= 1 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] -10 &= 1(20) + b \\[1pt] b &= -30 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_2 &= mt + b \\[1pt] a_2 &= (1)t - 30 \\[1pt] a_2 &= \left(t - 30\right)\ m/s^2 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} a_2 &= t - 30 \\[1pt] \frac{dv}{dt} &= t - 30 \\[1pt] \int_{300}^{v_2} dv &= \int_{10}^t \left(t - 30\right)\ dt \\[1pt] v_2 - 300 &= \left[\frac{t^2}{2} - 30t\right]_{10}^t \\[1pt] v_2 &= \left(\frac{t^2}{2} - 30t + 550\right)\ m/s \\[1pt] \end{aligned}

Displacement at `t = 10`

\begin{aligned} \left(S_2\right)_{initial} &= S_1(10) \\[1pt] \left(S_2\right)_{initial} &= 914.4\ m \\[1pt] \end{aligned}

Displacement function

\begin{aligned} v_2 &= \frac{t^2}{2} - 30t + 550 \\[1pt] \frac{dS}{dt} &= \frac{t^2}{2} - 30t + 550 \\[1pt] \int_{\left(S_2\right)_{initial}}^{S_2} dS &= \int_{10}^t \left(\frac{t^2}{2} - 30t + 550\right)\ dt \\[1pt] S_2 - 914.4 &= \left[\frac{t^3}{6} - 15t^2 + 550t\right]_{10}^t \\[1pt] S_2 &= \frac{t^3}{6} - 15t^2 + 550t - \frac{10^3}{6} + 15(10)^2 - 550(10) + 914.4 \\[1pt] S_2 &= \left(\frac{t^3}{6} - 15t^2 + 550t -3252.27\right)\ m \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_3 &= -10\ m/s^2 \\[1pt] \end{aligned}

Initial velocity of this interval

\begin{aligned} \left(v_3\right)_{initial} &= v_2(20) \\[1pt] \left(v_3\right)_{initial} &= 150\ m/s \\[1pt] \end{aligned}

Velocity function

\begin{aligned} a_3 &= -10 \\[1pt] \frac{dv}{dt} &= -10 \\[1pt] \int_{\left(v_3\right)_{initial}}^{v_3} dv &= \int_{20}^t -10\ dt \\[1pt] v_3 - 150 &= \Big[-10t\Big]_{20}^t \\[1pt] v_3 &= -10t + 200 + 150 \\[1pt] v_3 &= \left(-10t + 350\right)\ m/s \\[1pt] \end{aligned}

Initial displacement of this interval

\begin{aligned} \left(S_3\right)_{initial} &= S_2(20) \\[1pt] \left(S_3\right)_{initial} &= 3081.07\ m \\[1pt] \end{aligned}

Displacement function

\begin{aligned} v_3 &= \left(-10t + 350\right)\ m/s \\[1pt] \frac{dS}{dt} &= -10t + 350 \\[1pt] \int_{\left(S_3\right)_{initial}}^{S_3} dS &= \int_{20}^{t} \left(-10t + 350\right)\ dt \\[1pt] S_3 - 3081.07 &= \left[-5t^2 + 350t\right]_{20}^{t} \\[1pt] S_3 &= -5t^2 + 350t + 5(20)^2 - 350(20) + 3081.07 \\[1pt] S_3 &= -5t^2 + 350t -1918.93 \\[1pt] \end{aligned}

The jet will come to a stop when `v_3 = 0`.

\begin{aligned} v_3 &= -10t + 350 \\[1pt] 0 &= -10t + 350 \\[1pt] t &= 35s = t' \\[1pt] \end{aligned}

The velocity of a car is plotted as shown.

Determine the total distance the car moves until it stops `\left(t = 80\ s\right)`.

Construct the `a–t` graph.

Velocity function

\begin{aligned} v_1 &= 10 \\[1pt] \end{aligned}

Displacement function

\begin{aligned} S_1 &= \int_0^t 10\ dt \\[1pt] S_1 &= 10t \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_1 &= \frac{d}{dt} \left[10\right] \\[1pt] a_1 &= 0 \\[1pt] \end{aligned}

Slope

\begin{aligned} m &= \frac{\Delta v}{\Delta t} \\[1pt] m &= \frac{0 - 10}{80 - 40} \\[1pt] m &= -0.25 \\[1pt] \end{aligned}

`y` intercept

\begin{aligned} y &= mx + b \\[1pt] 0 &= -0.25(80) + b \\[1pt] b &= 20 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_2 &= mt + b \\[1pt] v_2 &= -0.25 t + 20 \\[1pt] \end{aligned}

Displacement function

\begin{aligned} S_2 &= \int_0^t \left(-0.25 t + 20\right)\ dt \\[1pt] S_2 &= -0.125 t^2 + 20 t \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_2 &= slope = m \\[1pt] a_2 &= -0.25 \\[1pt] \end{aligned}

The elevator starts from rest at the first floor of the building.

It can accelerate at `5\ ft/s^2` and then decelerate at `2\ ft/s^2`.

Determine the shortest time it takes to reach a floor `40\ ft` above the ground.

The elevator starts from rest and then stops.

Draw the `a–t, v–t` and `S–t` graphs for the motion.

Functions for when the elevator is accelerating.

Acceleration function

\begin{aligned} a_1 &= 5 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v &= u + at \\[1pt] v_1 &= 0 + (5)t \\[1pt] v_1 &= 5t \\[1pt] \end{aligned}

Position function

\begin{aligned} S &= ut + \frac{1}{2}at^2 \\[1pt] S_1 &= (0)t + \frac{1}{2}(5)t^2 \\[1pt] S_1 &= 2.5t^2 \\[1pt] \end{aligned}

Let's say `t_d` is the time the elevator starts decelerating.

Therefore, the initial velocity of the elevator when it starts decelerating is `v_1(t_d) = 5t_d`.

Express `t` in terms of `t_d` using the equation of motion evaluating for when the elevator is decelerating

\begin{aligned} v &= u + at \\[1pt] 0 &= 5t_d - 2\left(t - t_d\right) \color{green}\text{ // substituting deceleration values}\\[1pt] 0 &= 5t_d - 2t + 2t_d \\[1pt] 0 &= 7t_d - 2t \\[1pt] t &= \frac{7}{2}t_d \\[1pt] t &= 3.5t_d \\[1pt] \end{aligned}

Position function for when the elevator is decelerating

\begin{aligned} S &= ut + \frac{1}{2}at^2 \\[1pt] S_2 &= \left(5t_d\right)\left(t - t_d\right) + \frac{1}{2}(-2)\left(t - t_d\right)^2 \\[1pt] S_2 &= 5t_d\left(t - t_d\right) - \left(t - t_d\right)^2 \color{green}\text{ //position function 1} \\[1pt] \end{aligned}

Given that the total displacement is `40\ ft`, solve for `t_d`.

\begin{aligned} S_1(t_d) + S_2(t) &= 40 \\[1pt] 2.5 \left(t_d\right)^2 + 5t_d\left(t - t_d\right) - \left(t - t_d\right)^2 &= 40 \\[1pt] 2.5 \left(t_d\right)^2 + 5t_d\left(3.5t_d - t_d\right) - \left(3.5t_d - t_d\right)^2 &= 40 \color{green}\text{ //substituted } t = 3.5t_d\\[1pt] 8.75\left(t_d\right)^2 &= 40 \\[1pt] t_d &= 2.14\ s \\[1pt] \end{aligned}

The elevator is decelerating for `\displaystyle{\boxed{2.14\ seconds}}`.

\begin{aligned} t &= 3.5t_d \\[1pt] t &= 3.5 \left(2.14\right) \color{green}\text{ //substituted } t_d = 2.14\ s \\[1pt] t &= 7.48\ s \\[1pt] \end{aligned}

The shortest time it takes to reach `40\ ft` is `\displaystyle{\boxed{7.48\ seconds}}`.

Functions for when the elevator is decelerating.

Acceleration function

\begin{aligned} a_2 &= -2 \\[1pt] \end{aligned}

Position function

\begin{aligned} S_2 &= 5t_d\left(t - t_d\right) - \left(t - t_d\right)^2 \color{green}\text{ // position function 1} \\[1pt] S_2 &= 5(2.14)\left(t - 2.14\right) - \left(t - 2.14\right)^2 \color{green}\text{ //position function 2} \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v &= u + at \\[1pt] v_2 &= v_1(t_d) + (-2)t \\[1pt] v_2 &= 10.69 - 2(t - 2.14) \\[1pt] \end{aligned}

An airplane starts from rest, travels `5000\ ft` down a runway, and after uniform acceleration, takes off with a speed of `162\ mi/h`.

It then climbs in a straight line with a uniform acceleration of `3\ ft/s^2` until it reaches a constant speed of `220\ mi/h`.

Draw the `S–t`, `v–t` and `a–t` graphs that describe the motion.

Functions that describe the motion on the runway.

Acceleration function

\begin{aligned} v^2 &= u^2 + 2aS \\[1pt] 237.6^2 &= 0^2 + 2a(5000) \\[1pt] a_{runway} &= \frac{237.6^2}{10000} \\[1pt] a_{runway} &= 5.65\ ft/s^2 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v &= \int_0^t a\ dt \\[1pt] v_{runway} &= \int_0^t 5.65\ dt \\[1pt] v_{runway} &= 5.65t \\[1pt] \end{aligned}

Displacement function

\begin{aligned} S &= \int_0^t v\ dt \\[1pt] S_{runway} &= \int_0^t 5.65t\ dt \\[1pt] S_{runway} &= 2.82t^2 \\[1pt] \end{aligned}

`t_{take–off}` is the time on the runway or time until take off.

\begin{aligned} v &= u + at \\[1pt] v_{take–off} &= 0 + \left(a_{runway}\right) \left(t_{take–off}\right) \\[1pt] 237.6 &= 0 + \left(5.65\right) \left(t_{take–off}\right) \\[1pt] t_{take–off} &= 42.09s \\[1pt] \end{aligned}

Functions that describe the motion after the take off/during the climb.

Acceleration function

\begin{aligned} a_{climb} &= 3\ ft/s^2 \\[1pt] \end{aligned}

Displacement function

\begin{aligned} S &= ut + \frac{1}{2}at^2 \\[1pt] S_{climb} &= \left(v_{take–off}\right) t + \frac{1}{2}(3)t^2 \\[1pt] S_{climb} &= 237.6(t - 42.09) + 1.5(t - 42.09)^2 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v &= u + at \\[1pt] v_{climb} &= v_{take–off} + a \left(t - t_{take–off}\right) \\[1pt] v_{climb} &= 237.6 + 3(t - 42.09) \\[1pt] \end{aligned}

`t_{climb}` is the time till acceleration is active (post take off/during the climb), until constant speed is achieved.

\begin{aligned} v &= u + at \\[1pt] v_{climb} &= v_{take–off} + (3)t_{climb} \\[1pt] 322.67 &= 237.6 + (3)t_{climb} \\[1pt] t_{climb} &= 28.36 s \\[1pt] \end{aligned}

Total distance travelled

\begin{aligned} \Delta S &= S_{runway}(t_{take–off}) + S_{climb}(t_{take–off} + t_{climb}) \\[1pt] \Delta S &= 5000 + 7943.34 \\[1pt] \Delta S &= 12943.34\ ft \\[1pt] \end{aligned}

`t_{total}` is the total time the airplane was in motion till constant speed is achieved during the climb.

\begin{aligned} t_{total} &= t_{take–off} + t_{climb} \\[1pt] t_{total} &= 42.09 + 28.36 \\[1pt] t_{total} &= 70.44s \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v &= \frac{dS}{dt} \\[1pt] v &= \frac{d}{dt} \left[2\sin \left(\frac{\pi t}{5}\right) + 4\right] \\[1pt] v &= \frac{2\pi}{5}\cos \left(\frac{\pi t}{5}\right) \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a &= \frac{dv}{dt} \\[1pt] a &= \frac{d}{dt} \left[\frac{2\pi}{5}\cos \left(\frac{\pi t}{5}\right)\right] \\[1pt] a &= \frac{-2\pi^2}{25}\sin \left(\frac{\pi t}{5}\right) \\[1pt] \end{aligned}

A particle starts from `S = 0` and travels along a straight line with a velocity `v = \left(t^2 - 4t + 3\right)\ m/s`, where `t` is in seconds.

Construct the `v–t` and `a–t` graphs for the time interval `0 \le t \le 4s`.

Two rockets start from rest at the same elevation.

Rocket `A` accelerates vertically at `20\ m/s^2` for `12 s` and then maintains a constant speed.

Rocket `B` accelerates at `15\ m/s^2` until reaching a constant speed of `150\ m/s`.

Construct the `a–t, v–t` and `S–t` graphs for each rocket until `t = 20\ s`.

What is the distance between the rockets when `t = 20\ s`?

Acceleration function

\begin{aligned} a_{A_1} &= 20 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_{A_1} &= u + at \\[1pt] v_{A_1} &= 0 + (20)t \\[1pt] v_{A_1} &= 20t \\[1pt] \end{aligned}

Position function

\begin{aligned} S_{A_1} &= ut + \frac{1}{2}at^2 \\[1pt] S_{A_1} &= (0)t + \frac{1}{2}(20)t^2 \\[1pt] S_{A_1} &= 10t^2 \\[1pt] \end{aligned}

Final velocity when `t = 12s`

\begin{aligned} v_{A_1}(12) &= 20 \times 12 \\[1pt] v_{A_1}(12) &= 240 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_{A_2} &= 0 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_{A_2} &= u + at \\[1pt] v_{A_2} &= v_{A_1}(12) + (0)t \\[1pt] v_{A_2} &= 240 \\[1pt] \end{aligned}

Position function

\begin{aligned} S_{A_2} &= ut + \frac{1}{2}at^2 \\[1pt] S_{A_2} &= 240(t - 12) + \frac{1}{2}(0)(t - 12)^2 \\[1pt] S_{A_2} &= 240t - 2880 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_{B_1} &= 15 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_{B_1} &= u + at \\[1pt] v_{B_1} &= 0 + (15)t \\[1pt] v_{B_1} &= 15t \\[1pt] \end{aligned}

Position function

\begin{aligned} S_{B_1} &= ut + \frac{1}{2}at^2 \\[1pt] S_{B_1} &= (0)t + \frac{1}{2}(15)t^2 \\[1pt] S_{B_1} &= 7.5t^2 \\[1pt] \end{aligned}

Acceleration function

\begin{aligned} a_{B_2} &= 0 \\[1pt] \end{aligned}

Velocity function

\begin{aligned} v_{B_2} &= u + at \\[1pt] v_{B_2} &= v_{B_1}(10) + (0)t \\[1pt] v_{B_2} &= 150 \\[1pt] \end{aligned}

Position function

\begin{aligned} S_{B_2} &= ut + \frac{1}{2}at^2 \\[1pt] S_{B_2} &= (150)(t - 10) + \frac{1}{2}(0)(t - 10)^2 \\[1pt] S_{B_2} &= 150t - 1500 \\[1pt] \end{aligned}

Distance between the rockets

\begin{aligned} \Delta S &= \left(\text{Final position of Rocket A}\right) - \left(\text{Final position of Rocket B}\right) \\[1pt] \Delta S &= \left(S_{A_1}(12) + S_{A_2}(20)\right) - \left(S_{B_1}(10) + S_{B_2}(20)\right) \\[1pt] \Delta S &= \left(1440 + 1920\right) - \left(750 + 1500\right) \\[1pt] \Delta S &= 3360 - 2250 \\[1pt] \Delta S &= 1110\ m = 1.11\ km \\[1pt] \end{aligned}

The `S–t` graph for a train has been experimentally determined.

From the data, construct the `v–t` and `a–t` graphs for the motion; `0 \le t \le 40\ s`.

For `0 \le t \le 30 \ s `, the curve is `S = \left(0.4t^2\right)\ m`, and then it becomes straight for `t \ge 30\ s`.

A freight train starts from rest and travels with a constant acceleration of `0.5\ ft/s^2`.

After a time `t'` it maintains a constant speed so that when `t = 160\ s` it has traveled `2000\ ft`.

Determine the time `t'` and draw the `v–t` graph for the motion.

Functions for when the acceleration is constant at `0.5\ ft/s^2`

\begin{aligned} a_1 &= 0.5 \\[1pt] v_1 &= \int_0^t 0.5\ dt = 0.5t \\[1pt] S_1 &= \int_0^t 0.5t\ dt = 0.25t^2 \\[1pt] \end{aligned}

Total distance covered

\begin{aligned} S_1 &= 0.25t^2 \\[1pt] S_1(t') &= 0.25t'^2 \\[1pt] \end{aligned}

Position function when speed is constant i.e. acceleration is `zero`

The initial velocity `u` here is the final velocity of the first interval i.e. `v_1(t') = 0.5t'`

\begin{aligned} S_2 &= ut + \frac{1}{2}at^2 \\[1pt] S_2 &= (0.5t')t + \frac{1}{2}(0)t^2 \\[1pt] S_2 &= 0.5t't \\[1pt] \end{aligned}

Total distance covered

\begin{aligned} S_2(160 - t') &= (0.5t')(160 - t') \\[1pt] S_2(160 - t') &= 80t' - 0.5t'^2 \\[1pt] \end{aligned}

Solve for total distance travelled to find `t'`

\begin{aligned} S_{total} &= S_1(t') + S_2(160 - t') \\[1pt] 2000 &= 0.25t'^2 + 80t' - 0.5t'^2 \\[1pt] 2000 &= -0.25t'^2 + 80t' \\[1pt] 0.25t'^2 - 80t' + 2000 &= 0 \\[1pt] \hline t' &= 27.34\ s\ \textcolor{lime}{\checkmark} \color{green}\text{ //select this so } t \text{ is within } 160s \\[1pt] t' &= 292.66\ s \\[1pt] \end{aligned}

The constant velocity `u` in the second interval is the final velocity of the first interval i.e. velocity at `t'`

\begin{aligned} v_1 &= 0.5t \\[1pt] v_1(t') &= 0.5t' \\[1pt] v_1(27.34) &= 13.67\ ft/s \\[1pt] \end{aligned}

Graph

The dragster starts from rest and has a velocity described by the graph.

Construct the `S–t` graph during the time interval `0 \le t \le 15\ s`.

Also, determine the total distance traveled during this time interval.

Position when `t = 5`

\begin{aligned} S_1 &= 15t^2 \\[1pt] S_1(5) &= 15(5)^2 \\[1pt] S_1(5) &= 375\ m \\[1pt] \end{aligned}

Velocity function for the second interval

\begin{aligned} v &= -15t + 225 \\[1pt] \int_{375}^S dS &= \int_5^t \left(-15t + 225\right)\ dt \\[1pt] S_2 - 375 &= \left[-7.5t^2 + 225t\right]_5^t \\[1pt] S_2 - 375 &= -7.5t^2 + 225t + 7.5(5)^2 - 225(5) \\[1pt] S_2 &= -7.5t^2 + 225t -562.5 \\[1pt] \end{aligned}

The dragster starts from rest and has an acceleration described by the graph.

Construct the `v–t` graph for the time interval `0 \le t \le t'` where `t'` is the time for the car to come to rest.

Velocity at `t = 5`

\begin{aligned} v &= 20t, t \in [0, 5] \color{green}\text{ //first interval} \\[1pt] v(5) &= 100 \\[1pt] \end{aligned}

Velocity function for second interval

\begin{aligned} a &= -10 \\[1pt] \frac{dv}{dt} &= -10 \\[1pt] \int_{100}^{v} dv &= \int_5^{t} -10\ dt \\[1pt] v - 100 &= -10t + 50 \\[1pt] v &= 150 - 10t \color{green}\text{ //second interval} \\[1pt] \end{aligned}

Time `t'` when the car comes to rest

\begin{aligned} 0 &= 150 - 10t \color{green}\text{ //second interval} \\[1pt] t &= 15\ seconds = t' \\[1pt] \end{aligned}

The sports car travels along a straight road such that its acceleration is described by the graph.

Construct the `v–S` graph for the same interval and specify the velocity of the car when `S = 10\ m` and `S = 15\ m`.

A bicycle travels along a straight road where its velocity is described by the `v–S` graph.

Construct the `a–S` graph for the same interval.

A van travels along a straight road with a velocity described by the graph.

Construct the `S–t` and `a–t` graphs during the same period.

Take `S = 0` when `t = 0`.

The particle travels along a straight track such that its position is described by the `S–t` graph.

Construct the `v–t` graph for the same time interval.

Velocity when `x \in [0, 6]`

\begin{aligned} v_a &= \frac{d}{dt} \left[0.5t^3\right] \\[1pt] v_a &= 1.5t^2 \\[1pt] \end{aligned}

Velocity when `x \in (6, 10]`

\begin{aligned} v_b &= \frac{d}{dt} \left[108\right] \\[1pt] v_b &= 0 \\[1pt] \end{aligned}