Chapter 2, Problem 5SP

(a)

To determine

The distance travelled by each car at various times.

Answer to Problem 5SP

The distances travelled by the car A at each consecutive seconds are $2.25\text{m}$, $9\text{m}$, $20.25\text{m}$ and $36\text{m}$ respectively and for car B are $10\text{m}$, $20\text{m}$, $30\text{m}$ and $40\text{m}$ respectively.

Explanation of Solution

Given info: Velocity of the car B is $10\text{m/s}$ and acceleration of car A is $4.5{\text{m/s}}^{\text{2}}$.

Write the expression for distance.

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

Here,

$s$ is the distance

$v_0$ is the initial velocity

$t$ is the initial time

$a$ is the acceleration

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.

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

At 1s,

Substitute 1s for t and $4.5{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.5{\text{m/s}}^{\text{2}}\right){\left(1\text{s}\right)}^2\\ =2.25\text{m}\end{array}$

At 2s,

Substitute 2s for t and $4.5{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.5{\text{m/s}}^{\text{2}}\right){\left(2\text{s}\right)}^2\\ =9\text{m}\end{array}$

At 3s,

Substitute 3s for t and $4.5{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.5{\text{m/s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =20.25\text{m}\end{array}$

At 4s,

Substitute 4s for t and $4.5{\text{m/s}}^{\text{2}}$ for a to get s.

$\begin{array}{c}s=\frac{1}{2}\left(4.5{\text{m/s}}^{\text{2}}\right){\left(4\text{s}\right)}^2\\ =36\text{m}\end{array}$

For Car B,

Write the expression for distance.

$s=v_{0}t$

At 1s,

Substitute 1s for t and $10\text{m/s}$ for $v_0$ to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(1\text{s}\right)\\ =10\text{m}\end{array}$

At 2s,

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

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(2\text{s}\right)\\ =20\text{m}\end{array}$

At 3s,

Substitute 3s for t and $10\text{m/s}$ for $v_0$ to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(3\text{s}\right)\\ =30\text{m}\end{array}$

At 4s,

Substitute 4s for t and $10\text{m/s}$ for $v_0$ to get s.

$\begin{array}{c}s=\left(10\text{m/s}\right)\left(4\text{s}\right)\\ =40\text{m}\end{array}$

Conclusion:

Therefore, the distances travelled by the car A at each consecutive seconds are $2.25\text{m}$, $9\text{m}$, $20.25\text{m}$ and $36\text{m}$ respectively and for car B are $10\text{m}$, $20\text{m}$, $30\text{m}$ and $40\text{m}$ respectively.

(b)

To determine

The time when Car A overtakes Car B.

Answer to Problem 5SP

Car A overtakes Car B between the interval $4\text{ s}$ and $5\text{s}$.

Explanation of Solution

Car A will overtake car B only when the distance travelled by car A becomes more than the distance travelled by car B. The result in part (a) shows that this does not happen till $4\text{ s}$ but the difference in the distances is getting smaller. This means that the car A will overtake car B after $4\text{ s}$ but before reaching $5\text{ s}$.

Conclusion:

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

(c)

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 $4.44\text{s}$.

Explanation of Solution

Given info: Velocity of the car B is $10\text{m/s}$ and acceleration of car A is $4.5{\text{m/s}}^{\text{2}}$.

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 $10\text{m/s}$ for v and $4.5{\text{m/s}}^{\text{2}}$ for a to get t.

$\begin{array}{c}t=\frac{2\left(10\text{m/s}\right)}{4.5{\text{m/s}}^{\text{2}}}\\ =4.44\text{s}\end{array}$

Conclusion:

Therefore, the time at which car A overtakes car B is $4.44\text{s}$.