A car is traveling at a constant velocity of 21 m/s (about 45 mph). The driver s

Question

A car is traveling at a constant velocity of 21 m/s (about 45 mph). The driver sees a fallen tree lying across the road and hits the brakes after a delay (reaction time) of 240 ms. (1 ms = 10-3 s.)

A) Calculate the distance the car travels during this 240 ms reaction time period.

HINT: This part is straightforward, but you need to understand the setup. During the "reaction time", the car continues to travel at the stated constant velocity. It is only after the reaction time that the brakes are applied. That's the next part of the problem.

B) Once the brakes are applied, the car slows to a stop with an acceleration of -5.28 m/s2. Calculate the distance the car travels while the brakes are applied.

HINTS: You should find that you know the initial and final velocities during the time the brakes are applied, and you have been given the acceleration. You want to find the displacement, or the distance that the car travels. There are two ways to solve this problem. One is a two-step approach in which you first find the time, and then find the displacement. Another is a one-step approach where you use that one uniform acceleration equation that I tend to forget (but which you may find in the book or from our worksheet).

Explanation / Answer

(A) In reaction time he did not applied break and moving with the same speed

V= 21 m/s and time t = 240 x 10-3 sec

so distance covered when an object move with constant speed is given by

D = V*t

D = (21 m/s)(240 x 10-3)

D = 5.04 m

(B) Final velocity Vf = 0

acceleration a = -5.28 m/s2

Vf2 = V2 + 2ad

0 = (21)2 + 2(-5.28)(d)

d = 41.76 m

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