Question

FIGURE Q2.14 shows the velocity-versus-time graph for a moving object. At which lettered point or points:

a. Is the object speeding up?

b. Is the object slowing down?

c. Is the object moving to the left?

d. Is the object moving to the right?

Short answer

a. The object is speeding up at the point B.

b. The object is slowing down at the point A.

c. The object is moving to the left at the point A.

d. The object is moving to the right at the point C.

Step-by-step solution

Part a Step 1: Introduction

The tangent at a point of the position-time curve shows the direction of speed at that point.

Explanation

At point B, the slope of speed-time graph is increasing. So at point B, the object is speeding up.

Part b Step 1: Introduction

When the direction of acceleration is considered as negative then the object will be slowing down.

Explanation

At point A, the direction of acceleration of the object is towards negative. So the magnitude of the speed at this point should be slowed down.

Part C Step 1: Introduction

The slope of the graph with positive x-axis determines the velocity of the object. If the slope is negative then the direction of the object is also considered as opposite compare to the first condition.

Explanation

At point A, the object is accelerating in negative direction. The slope of the graph is in negative direction at this point. So at point A, the object is moving to the left.

Part d Step 1: Introduction

If the slope of x-t graph is positive, then the acceleration of the object should be positive in magnitude and direction.

Explanation

At point C of the given graph, the slope of the graph is positive and the magnitude of acceleration of the object is positive. So the object is moving forward with an increasing speed or moving to the right direction.