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Chapter 2: Q. 14 (page 58)

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

Expert verified

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

01

Part a Step 1: Introduction

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

02

Explanation

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

03

Part b Step 1: Introduction

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

04

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.

05

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.

06

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.

07

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.

08

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.

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Most popular questions from this chapter

Careful measurements have been made of Olympic sprinters

in the 100 meter dash. A quite realistic model is that the sprinter鈥檚

velocity is given by

vx=a(1-e-bt)

where t is in s, vx is in m/s, and the constants a and b are characteristic

of the sprinter. Sprinter Carl Lewis鈥檚 run at the 1987

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b = 0.6887 s-1.

a. What was Lewis鈥檚 acceleration at t = 0 s, 2.00 s, and 4.00 s?

b. Find an expression for the distance traveled at time t.

c. Your expression from part b is a transcendental equation,

meaning that you can鈥檛 solve it for t. However, it鈥檚 not hard to

use trial and error to find the time needed to travel a specific

distance. To the nearest 0.01 s, find the time Lewis needed to

sprint 100.0 m. His official time was 0.01 s more than your

answer, showing that this model is very good, but not perfect.

A particle鈥檚 velocity is described by the function vx=t2-7t+10ms, where t is in s. a. At what times does the particle reach its turning points? b. What is the particle鈥檚 acceleration at each of the turning points?

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes.

a. How far does Tina drive before passing David?

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FIGURE P2.46 shows a set of kinematic graphs for a ball rolling on a track. All segments of the track are straight lines, but some may be tilted. Draw a picture of the track and also indicate the ball鈥檚 initial condition.

Particles A, B, and C move along the x-axis. Particle C has an initial velocity of 10 m/s. In FIGURE P2.37, the graph for A is a position-versus-time graph; the graph for B is a velocity-versus time graph; the graph for C is an acceleration-versus-time graph. Find each particle鈥檚 velocity at t = 7.0 s.
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