A straight road is used to test the driver\'s reaction time and the car\'s accel

Question

A straight road is used to test the driver's reaction time and the car's acceleration. The driver is instructed to immediately push the brake pedal
when the car is just passing a white marker post and to keep pressing the brake pedal until the car comes to a complete stop. The distance from the car
to the white post is then measured. The driver's reaction time is the time between when the car just passes the post and when the brake pedal is pushed,
The driver's reaction time and the car's acceleration were the same for both runs.
The 1200 kg car is 4.8 m long. The 50 kg driver is 2.2 m tall.

For the first run, the car was moving at 16.5 when it passed the white post and stopped 28.7 m from the post.
For the second run, the car was moving at 28.4 when it passed the white post and stopped 74.7 m from the post.


a) What was the magnitude of the car's acceleration after the brake pedal was pushed?

1

b) What was the driver's reaction time?

2 s

c) For the first run, how much time did it take for the car to stop after the brake pedal was pushed?

3 s

d) For the first run, how far did past the white post did the driver push the brake pedal?

4 m

e) For the first run, how much further did the car travel after the brake pedal was pushed?

5 m

Explanation / Answer

Hi, Let t1 be the reaction time of the driver. so if t is the total time taken for the car to stop in the first run from the point it passed the white post, then t = t1 + t2 where t2 is the time taken by the car to stop from the instant brakes are applied. similarly for the second run, t = t1 + t3 where t3 is the time taken by the car to stop from the instant brakes are applied. So in each run it travelled some distance with constant velocity for a time t1 and then brakes are applied. Hence the expressions for the total displacement in the first and second run becomes. s1 = (u1*t1) + [u1*t2 + (1/2)at2^2] and s2 = (u2*t1) + [u2*t3 + (1/2)at3^2] (just remember that time of reaction t1 and acceleration of the brakes are same in both cases). Hence from given data the above expressions become   95.1 = (26.4 * t1) + [(26.4 * t2) + (0.5 * a * t2^2)] and   251.6 = (45.4 * t1) + [(45.4 * t3) + (0.5 * a * t3^2)] Also from v = u + at, we have for both runs, 0 = 26.4 + a*t2 and 0 = 45.4 + a*t3. => t2 = -26.4/a and      t3 = -45.4/a. Substitute the values of t2 and t3 in the above equations to solve for t1 and a we get, a) acceleration = -4.949 m/s (-ve sign implies its decceleration), b)  t1 = reaction time = 0.935 sec c) t2 = -26.4/a = 5.334 sec
d)  x = (26.4 * t1) = 24.684 m
e) x' = s1 - x = 95.1 - 24.684 = 70.416 m

Hope this helps you. Hi, Let t1 be the reaction time of the driver. so if t is the total time taken for the car to stop in the first run from the point it passed the white post, then t = t1 + t2 where t2 is the time taken by the car to stop from the instant brakes are applied. similarly for the second run, t = t1 + t3 where t3 is the time taken by the car to stop from the instant brakes are applied. So in each run it travelled some distance with constant velocity for a time t1 and then brakes are applied. Hence the expressions for the total displacement in the first and second run becomes. s1 = (u1*t1) + [u1*t2 + (1/2)at2^2] and s2 = (u2*t1) + [u2*t3 + (1/2)at3^2] (just remember that time of reaction t1 and acceleration of the brakes are same in both cases). Hence from given data the above expressions become   95.1 = (26.4 * t1) + [(26.4 * t2) + (0.5 * a * t2^2)] and   251.6 = (45.4 * t1) + [(45.4 * t3) + (0.5 * a * t3^2)] Also from v = u + at, we have for both runs, 0 = 26.4 + a*t2 and 0 = 45.4 + a*t3. => t2 = -26.4/a and      t3 = -45.4/a. Substitute the values of t2 and t3 in the above equations to solve for t1 and a we get, a) acceleration = -4.949 m/s (-ve sign implies its decceleration), b)  t1 = reaction time = 0.935 sec c) t2 = -26.4/a = 5.334 sec
d)  x = (26.4 * t1) = 24.684 m
e) x' = s1 - x = 95.1 - 24.684 = 70.416 m

Hope this helps you.

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