Chapter 3, Problem 67A

a.

To determine

Distance travelled by car before it stops.

Answer to Problem 67A

After travelling 137.5 m car will stop.

Explanation of Solution

Given :

Acceleration of car is $a=-11{\text{ m/sec}}^2$ (opposite direction of motion)

Initial velocity of car is $u=55\text{ m/sec}$

Formula used:

According to third equation of motion.

  $v^2=u^2+2as$

Where,

a is acceleration

v is final velocity

s is distance travelled.

u is initial velocity

Calculation:

As the car finally stops, therefore final velocity of car is $v=0\text{ m/sec}$

Substitute the value of u, v and a.

  $\begin{array}{c}{0}^2={55}^2+2\times \left(-11\right)\times s\\ 22s={55}^2\\ s=\frac{3025}{22}\\ =137.5\text{ m}\end{array}$

Conclusion:

Therefore, after travelling 137.5 m car will stop.

b.

To determine

Distance travelled by car before it stops.

Answer to Problem 67A

After travelling 550 m car will stop.

Explanation of Solution

Given :

Acceleration of car is given $a=-11{\text{ m/sec}}^2$ ( opposite direction of motion)

Initial velocity of car is twice of that in part (a).

Formula used:

According to third equation of motion.

  $v^2=u^2+2as$

Where,

a is acceleration

v is final velocity

s is distance travelled.

u is initial velocity

Calculation:

As the car finally stops therefore final velocity of car is $v=0\text{ m/sec}$

The initial velocity is,

  $\begin{array}{c}u=2\times 55\\ =110\text{ m/sec}\end{array}$

Substitute the value of u, v and a.

  $\begin{array}{c}{0}^2={110}^2+2\times \left(-11\right)\times s\\ 22s={110}^2\\ s=\frac{12100}{22}\\ =550\text{ m}\end{array}$

Conclusion:

Therefore, after travelling 550 m car will stop.