Chapter 2, Problem 4SP

A car traveling in a straight line with an initial velocity of 10 m/s accelerates at a rate of 3.0 m/s² to a velocity of 34 m/s.

1. a. How much time does it take for the car to reach the velocity of 34 m/s?
2. b. What is the distance covered by the car in this process?
3. c. Compute values of the distance traveled at 1-second intervals and carefully draw a graph of distance plotted against time for this motion.

To determine

The time taken for the car to reach the velocity of 34 m/s.

Answer to Problem 4SP

The time taken for the car to reach the velocity of 34 m/s is $8\text{s}$.

Explanation of Solution

Given info: Initial velocity of the car is 10 m/s, acceleration is $3.0{\text{m/s}}^{\text{2}}$ and final velocity is 34 m/s.

Write the expression for Newton’s equation of motion.

$v=v_0+at$

Here,

v is the final velocity

$v_0$ is the initial velocity

t is the initial time

a is the acceleration

Re-arrange the above equation to get t.

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

Substitute 34 m/s for v, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get t.

$\begin{array}{c}t=\frac{\left(34\text{m/s}\right)-\left(10\text{m/s}\right)}{3.0{\text{m/s}}^{\text{2}}}\\ =8\text{s}\end{array}$

Conclusion:

The time taken for the car to reach the velocity of 34 m/s is $8\text{s}$.

To determine

The distance covered by the the car.

Answer to Problem 4SP

The distance covered by the car is $176\text{m}$.

Explanation of Solution

Given info: Initial velocity of the car is 10 m/s, acceleration is $3.0{\text{m/s}}^{\text{2}}$ and final velocity is 34 m/s.

Write the expression for distance.

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

Here,

v is the final velocity

$v_0$ is the initial velocity

t is the initial time

s is the distance

Substitute 8s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(8\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(8\text{s}\right)}^2\\ =176\text{m}\end{array}$

The distance covered by the car is $176\text{m}$.

To determine

To plot: The graph of distance versus time.

Answer to Problem 4SP

Distance is plotted against time in the graph.

Explanation of Solution

Write the expression for distance.

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

At 0s,

Substitute 0s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(0\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(0\text{s}\right)}^2\\ =0\text{m}\end{array}$

At 1s,

Substitute 1s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(1\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\\ =11.5\text{m}\end{array}$

At 2s,

Substitute 2s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(2\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(2\text{s}\right)}^2\\ =26\text{m}\end{array}$

At 3s,

Substitute 3s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(3\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =43.5\text{m}\end{array}$

At 4s,

Substitute 4s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(4\text{s}\right)}^2\\ =64\text{m}\end{array}$

At 5s,

Substitute 5s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(5\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(5\text{s}\right)}^2\\ =87.5\text{m}\end{array}$

At 6s,

Substitute 6s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(6\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(6\text{s}\right)}^2\\ =114\text{m}\end{array}$

At 7s,

Substitute 7s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(7\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(7\text{s}\right)}^2\\ =143.5\text{m}\end{array}$

At 8s,

Substitute 8s for t, 10 m/s for $v_0$ and $3.0{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(8\text{s}\right)+\frac{1}{2}\left(3.0{\text{m/s}}^{\text{2}}\right){\left(8\text{s}\right)}^2\\ =176\text{m}\end{array}$

The above values are plotted in the below graph.

Conclusion:

Distance is plotted against time in the graph.