Essential University Physics (3rd Edition)
3rd Edition
ISBN: 9780134202709
Author: Richard Wolfson
Publisher: PEARSON
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Textbook Question
Chapter 2, Problem 96PP
A wildlife biologist is studying the hunting patterns of tigers. She anesthetizes a tiger and attaches a GPS collar to track its movements. The collar transmits data on the tiger’s position and velocity. Figure 2.16 shows the tiger’s velocity as a function of time as it moves on a one-dimensional path
FIGURE 2.16 The tiger’s velocity (Passage Problems 92-96)
At which point is the tiger farthest from its starting position at t = 0?
- a. C
- b. E
- c. F
- d. H
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Please help solve these questions, the solutions are given below I'm just not sure how to solve it.
(Use paper sheet ,Not Typewritten)
The solutions are:
1) 110
2)(sqrt(t+0.25))-0.5
3)1.00
My question isn't how to solve the problem exactly. In fact, it's already been solved on this website. My question is about the acceleration. When I solve this problem myself, first I calculate the velocity by dividing 100m by 53s. I get 1.89m/s. Then I use that to find the acceleration using the equation vf = vi + at. That's 1.89/53 = 0.036m/s^2.
That's not correct. The correct way to find the acceleration is to us the equation d = 1/2 at^2 and solve that way without taking the intermediate step of finding the velocity. Doing it that way, the acceleration is 0.0712m/s^2.
My question is why you get a different result doing it the first way than you get doing it the second way.
Chapter 2 Solutions
Essential University Physics (3rd Edition)
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