Chapter 3, Problem 65A

a.

To determine

Displacement of the car after 6 sec.

Answer to Problem 65A

Displacement of the car after 6 sec is 43.2 m.

Explanation of Solution

Given :

Acceleration of car is given $a=-1.6{\text{ m/sec}}^2$ (backward acceleration)

Initial velocity of car is $u=12\text{ m/sec}$

Time taken is $t=6\text{ sec}$

Formula used:

According to second equation of motion,

  $s=ut+\frac{1}{2}a{t}^2$

Where,

s is displacement

a is acceleration

u is initial velocity

t is time.

Calculation:

Substitute the value of u, a and tin equation of motion.

  $\begin{array}{c}s=12\times 6+\frac{1}{2}\times \left(-1.6\right)\times 6^2\\ =72-28.8\\ =43.2\text{ m}\end{array}$

Conclusion:

Therefore, displacement of the car after 6 sec is 43.2 m.

b.

To determine

Displacement of the car after 9 sec.

Answer to Problem 65A

Displacement of the car after 9 sec is 43.2 m

Explanation of Solution

Given :

Acceleration of car is given $a=-1.6{\text{ m/sec}}^2$ (backward acceleration)

Initial velocity of car is $u=12\text{ m/sec}$

Time taken is $t=9\text{ sec}$

Formula used:

According to second equation of motion,

  $s=ut+\frac{1}{2}a{t}^2$

Where,

s is displacement

a is acceleration

u is initial velocity

t is time.

Calculation:

Substitute the value of u, a and tin equation of motion.

  $\begin{array}{c}s=12\times 9+\frac{1}{2}\times \left(-1.6\right)\times 9^2\\ =108-64.8\\ =43.2\text{ m}\end{array}$

Conclusion:

Therefore, displacement of the car after 9 sec is 43.2 m.