Chapter 2, Problem 5SP

Just as car A is starting up, it is passed by car B. Car B travels with a constant velocity of 7 m/s, while car A accelerates with a constant acceleration of 4.2 m/s², starting from rest.

1. a. Compute the distance traveled by each car for times of 1 s, 2 s, 3 s, and 4 s.
2. b. At what time, approximately, does car A overtake car B?
3. c. How might you go about finding this time exactly? Explain.

To determine

The distance travelled by the car at various times.

Answer to Problem 5SP

The distances travelled by the car at various times are calculated using Newton’s equations of motion.

Explanation of Solution

Given info: Velocity of the car B is 7 m/s and acceleration of car A is $4.2{\text{m/s}}^{\text{2}}$.

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

Car A starts from rest. Hence, initial velocity is zero. Car B travels with constant velocity. Hence, its acceleration is zero.

For Car A,

Write the expression for distance.

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

Here,

t is the initial time

s is the distance

a is the acceleration

At 1s,

Substitute 1s for t and $4.2{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.2{\text{m/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\\ =2.1\text{m}\end{array}$

At 2s,

Substitute 2s for t and $4.2{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.2{\text{m/s}}^{\text{2}}\right){\left(2\text{s}\right)}^2\\ =8.4\text{m}\end{array}$

At 3s,

Substitute 3s for t and $4.2{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.2{\text{m/s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =18.9\text{m}\end{array}$

At 4s,

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

$\begin{array}{c}s=\frac{1}{2}\left(4.2{\text{m/s}}^{\text{2}}\right){\left(4\text{s}\right)}^2\\ =33.6\text{m}\end{array}$

For Car B,

Write the expression for distance.

$s=vt$

Here,

t is the initial time

s is the distance

v is the velocity

At 1s,

Substitute 1s for t and 7 m/s for v to get s.

$\begin{array}{c}s=\left(7\text{m/s}\right)\left(1\text{s}\right)\\ =7\text{m}\end{array}$

At 2s,

Substitute 2s for t and 7 m/s for v to get s.

$\begin{array}{c}s=\left(7\text{m/s}\right)\left(2\text{s}\right)\\ =14\text{m}\end{array}$

At 3s,

Substitute 3s for t and 7 m/s for v to get s.

$\begin{array}{c}s=\left(7\text{m/s}\right)\left(3\text{s}\right)\\ =21\text{m}\end{array}$

At 4s,

Substitute 4s for t and 7 m/s for v to get s.

$\begin{array}{c}s=\left(7\text{m/s}\right)\left(4\text{s}\right)\\ =28\text{m}\end{array}$

Conclusion:

The distances travelled by the car at various times are calculated using Newton’s equations of motion.

To determine

The time when Car A overtakes Car B.

Answer to Problem 5SP

Car A overtakes Car B between the interval 3s and 4s.

Explanation of Solution

From (a), it can be observed that the distance travelled by car A is more than that of car B at 4s. Hence, it is evident that somewhere between 3s and 4s, car A must have overtaken car B.

Conclusion:

Car A overtakes Car B between the interval 3s and 4s.

To determine

The time at which car A overtakes car B.

Answer to Problem 5SP

The time at which car A overtakes car B is $3.33\text{s}$.

Explanation of Solution

Given info: Velocity of the car B is 7 m/s and acceleration of car A is $4.2{\text{m/s}}^{\text{2}}$.

Equate the expressions for distances travelled by car A and car B.

$\frac{1}{2}a{t}^2=vt$

Re-arrange the above equation to get t.

$t=\frac{2v}{a}$

Substitute 7 m/s for v and $4.2{\text{m/s}}^{\text{2}}$ for a to get t.

$\begin{array}{c}t=\frac{2\left(7\text{m/s}\right)}{4.2{\text{m/s}}^{\text{2}}}\\ =3.33\text{s}\end{array}$

Conclusion:

The time at which car A overtakes car B is $3.33\text{s}$.