College Physics

Figure 2.22

Strategy

Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.

Solution

1. Identify the knowns.x′f= 3.75 km,x′ 0 = 5.25 km,Δt= 5.00 min.

2. Determine displacement,Δx′. We foundΔx′to be − 1.5 kminExample 2.2.

3. Solve for average velocity.
   (2.19)

v- =Δx′

Δt

=−1.50 km

5.00 min

4. Convert units.
   (2.20)

v

-

=Δx′

Δt

=

⎛

⎝

−1.50 km

5.00 min

⎞

⎠

⎛

⎝

60 min

1 h

⎞

⎠= −18.0 km/h

Discussion

The negative velocity indicates motion to the left.

Example 2.7 Calculating Deceleration: The Subway Train

Finally, suppose the train inFigure 2.22slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?

Strategy

Once again, let’s draw a sketch:

Figure 2.23

As before, we must find the change in velocity and the change in time to calculate average acceleration.

Solution

1. Identify the knowns.v 0 = −20 km/h,vf= 0 km/h,Δt= 10.0 s.

2. CalculateΔv. The change in velocity here is actually positive, since

Δv=vf−v 0 = 0 −(−20 km/h)=+20 km/h. (2.21)

3. Solve for a-.

(2.22)

a-=Δv

Δt

=+20.0 km/h

10.0 s

4. Convert units.
   (2.23)

a

-

=

⎛

⎝

+20.0 km/h

10.0 s

⎞

⎠

⎛

⎝

103 m

1 km

⎞

⎠

⎛

⎝

1 h

3600 s

⎞

⎠= +0.556 m/s

2

Discussion

The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this

problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as thechangein

50 CHAPTER 2 | KINEMATICS

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