Chapter 2, Problem 18E

Starting from rest, a car accelerates at a constant rate of 4.0 m/s² for a time of 5 seconds.

1. a. Compute the velocity of the car at 1 s, 2 s, 3 s, 4 s, and 5 s and plot these velocity values against time.
2. b. Compute the distance traveled by the car for these same times and plot the distance values against time.

To determine

The velocity of the car after each second.

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are 4 m/s, 8 m/s, 12 m/s, 16 m/s and 20 m/s respectively.

Explanation of Solution

Given info: Acceleration is $4{\text{m/s}}^{\text{2}}$ and time is 5 seconds.

Write the formula for average acceleration.

$a=\frac{v-v_0}{t}$

Here,

$v$ is the final velocity

$v_0$ is the initial velocity

a is the average acceleration

t is the time

Re-arrange the above equation to get $v_0$.

$v=v_0+at$

For $t=1\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 1 s for t to get v.

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(4{\text{m/s}}^{\text{2}}\right)\left(1\text{s}\right)\\ =4\text{m/s}\end{array}$

For $t=2\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 2 s for t to get v.

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(4{\text{m/s}}^{\text{2}}\right)\left(2\text{s}\right)\\ =8\text{m/s}\end{array}$

For $t=3\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 3 s for t to get v.

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(4{\text{m/s}}^{\text{2}}\right)\left(3\text{s}\right)\\ =12\text{m/s}\end{array}$

For $t=4\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 4 s for t to get v.

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(4{\text{m/s}}^{\text{2}}\right)\left(4\text{s}\right)\\ =16\text{m/s}\end{array}$

For $t=5\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 5 s for t to get v.

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(4{\text{m/s}}^{\text{2}}\right)\left(5\text{s}\right)\\ =20\text{m/s}\end{array}$

The velocity time graph is given below.

Conclusion:

The velocities after 1s, 2s, 3s, 4s and 5s are 4 m/s, 8 m/s, 12 m/s, 16 m/s and 20 m/s respectively.

To determine

The distance travelled after each second.

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are 2 m, 8 m, 18 m, 32 m and 50 m respectively.

Explanation of Solution

Given info: Acceleration is $4{\text{m/s}}^{\text{2}}$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v_{0}t+\frac{1}{2}a{t}^2$

Here,

d is the distance

For $t=1\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 1 s for t to get d.

$\begin{array}{c}d=\left(0\text{m/s}\right)\left(1\text{s}\right)+\frac{1}{2}\left(4{\text{m/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\\ =2\text{m}\end{array}$

For $t=2\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 2 s for t to get d.

$\begin{array}{c}d=\left(0\text{m/s}\right)\left(2\text{s}\right)+\frac{1}{2}\left(4{\text{m/s}}^{\text{2}}\right){\left(2\text{s}\right)}^2\\ =8\text{m}\end{array}$

For $t=3\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 3 s for t to get d.

$\begin{array}{c}d=\left(0\text{m/s}\right)\left(3\text{s}\right)+\frac{1}{2}\left(4{\text{m/s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =18\text{m}\end{array}$

For $t=4\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 4 s for t to get d.

$\begin{array}{c}d=\left(0\text{m/s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(4{\text{m/s}}^{\text{2}}\right){\left(4\text{s}\right)}^2\\ =32\text{m}\end{array}$

For $t=5\text{s}$,

Substitute 0 m/s for $v_0$, $4{\text{m/s}}^{\text{2}}$ for a and 5 s for t to get d.

$\begin{array}{c}d=\left(0\text{m/s}\right)\left(5\text{s}\right)+\frac{1}{2}\left(4{\text{m/s}}^{\text{2}}\right){\left(5\text{s}\right)}^2\\ =50\text{m}\end{array}$

The distance time graph is,

Conclusion:

The distances after 1s, 2s, 3s, 4s and 5s are 2 m, 8 m, 18 m, 32 m and 50 m respectively.