a car travels a distance d = 21.1m in the positive x direction in a time t1=22.7s, at which the point the car brakes, coming to a rest in t2=5.08s

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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a car travels a distance d = 21.1m in the positive x direction in a time t1=22.7s, at which the point the car brakes, coming to a rest in t2=5.08s

Expert Solution
Step 1

During the first part of the motion, the car is traveling with a constant velocity v1. We can find this velocity using the formula:

v1 = d / t1

Substituting the given values, we get:

v1 = 21.1 m / 22.7 s = 0.93 m/s

During the second part of the motion, the car is decelerating with a constant acceleration a, until it comes to a stop. We can find the acceleration using the formula:

a = (v_f - v_i) / t

where v_f is the final velocity (which is zero), v_i is the initial velocity (which is v1), and t is the time it takes to come to a stop (which is t2).

Substituting the given values, we get:

a = (0 - 0.93 m/s) / 5.08 s

= -0.183 m/s^2

So the deceleration of the car is 0.183 m/s^2.

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