Chapter 2, Problem 51P

(a)

To determine

The time period when the velocity is constant.

Answer to Problem 51P

Velocity is constant in $t=0\text{s}-20\mathrm{s}$

Explanation of Solution

Given:

The given graph is shown below.

  

Formula Used:

Velocity is calculated as $v=\frac{\Delta s}{\Delta t}$

Calculation:

Constant velocity is obtained by deducing the slope of the distance-time graph. Here, velocity is constant in the starting period of $t=0\text{s}-20\mathrm{s}$

Conclusion:

Velocity is constant in $t=0\text{s}-20\mathrm{s}$

(b)

To determine

The time when the velocity is greatest.

Answer to Problem 51P

Velocity is maximum at $t=30\text{ sec}$

Explanation of Solution

Given:

The given graph is shown below.

  

Calculation:

Velocity is maximum at its peak when the slope would be highest positive. For the given graph, the slope is highestat $t=30\sec$

Conclusion:

Velocity is maximum at $t=30\sec$

(c)

To determine

The time when the velocity is zero.

Answer to Problem 51P

Velocity will be zero at $t=37\sec$

Explanation of Solution

Given:

The given graph is shown below.

  

Calculation:

Velocity will be zero when the instantaneous position of the object will be zero that is when the graph is at the highest possible value at $t=37\sec$

Conclusion:

Velocity will be zero at $t=37\sec$

(d)

To determine

To identify: Whether the object will move in one direction or both directions during the time shown.

Answer to Problem 51P

The object will move forward in $t=0\text{s}-37\mathrm{s}$ and backward in $t=37\text{s}-46\mathrm{s}$

Explanation of Solution

Given:

The given graph is shown below.

  

Calculation:

As shown in the graph, the object moves in both positive and negative direction. When the slope is positively increasing then the object will move in the forward direction in $t=0\text{s}-37\mathrm{s}$ . But, when the slope is negatively decreasing then the object is moving in the backward direction at $t=37\text{s}-46\mathrm{s}$

Conclusion:

The object will move forward in $t=0\text{s}-37\mathrm{s}$ and backward in $t=37\text{s}-46\mathrm{s}$