Chapter 3, Problem 59A

a.

To determine

Acceleration of the object.

Answer to Problem 59A

Acceleration of object for first 5 min is $6{\text{ m/min}}^2$ .

Explanation of Solution

Given:

Velocity time graph is given

  

Time is from 0 to 5 min. So, t = 5 min

Formula Used:

Acceleration of an object is equal to change if velocity is with respect to time.

Mathematically acceleration is

  $a=\frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is the time taken to reach final velocity from initial velocity.

Calculation:

For the first 5 min, velocity changes from 0 m/min to 30 m/min.

Therefore, initial velocity u = 0 m/min

And final velocity v = 30 m/min.

So, on substituting the values

  $\begin{array}{c}a=\frac{30-0}{5}\\ =6{\text{ m/min}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the object is $a=6{\text{ m/min}}^2$ .

b.

To determine

Acceleration of the object.

Answer to Problem 59A

Acceleration of object for between 5 to 10 min is $0{\text{ m/min}}^2$ .

.

Explanation of Solution

Given:

Velocity time graph is given

  

Time is from 5 to 10 min. So, t = 5 min

Formula Used:

Acceleration of an object is equal to change if velocity is with respect to time.

Mathematically acceleration is

  $a=\frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is the time taken to reach final velocity from initial velocity.

Calculation:

Between 5 to 10 min, there is no change in velocity.

Therefore, initial velocity u = 30 m/min, and final velocity v = 30 m/min.

So, on substituting the values,

  $\begin{array}{c}a=\frac{30-30}{5}\\ =0{\text{ m/min}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the object is $a=0{\text{ m/min}}^2$ .

c.

To determine

Acceleration of the object.

Answer to Problem 59A

Acceleration of object for between 10 to 15 min is $-2{\text{ m/min}}^2$ .

Explanation of Solution

Given:

Velocity time graph is given

  

Time is from 10 to 15 min. So, t = 5 min.

Formula Used:

Acceleration of an object is equal to change if velocity with respect to time.

Mathematically acceleration is

  $a=\frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is the time taken to reach final velocity from initial velocity.

Calculation:

Between 10 to 15 min velocity changes from 30 m/min to 20 m/min.

Therefore, initial velocity u = 30 m/min.

And final velocity v = 20 m/min.

So, on substituting the values

  $\begin{array}{l}a=\frac{20-30}{5}\\ =-2{\text{ m/min}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the object is $a=-2{\text{ m/min}}^2$ .

d.

To determine

Acceleration of the object.

Answer to Problem 59A

Acceleration of object for between 20 to 25 min is $-4{\text{ m/min}}^2$ .

Explanation of Solution

Given:

Velocity time graph is given;

  

Time is from 20 to 25 min. So, t = 5 min

Formula Used:

Acceleration of an object is equal to change if velocity with respect to time.

Mathematically acceleration is

  $a=\frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is the time taken to reach final velocity from initial velocity.

Calculation:

Between 20 to 25 min, the velocity changes from 20 m/min to 0 m/min.

Therefore, initial velocity u = 20 m/min.

And final velocity v = 0 m/min.

So, on substituting the values

  $\begin{array}{c}a=\frac{0-20}{5}\\ =-4{\text{ m/min}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the object is $a=-4{\text{ m/min}}^2$ .