Question

You're driving down the highway late one night at 22 m/s when a deer steps onto the road 38 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 .

Answer 1

speed of the car is given on the road

$v = 22 m/s$

reaction time is given as

$t = 0.50 s$

now we can find the distance that it will move during this reaction time

$d_1 = 22* 0.50 = 11 m$

now the deceleration of the car is 10 m/s^2

so the distance that it will move before stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0^2 - 22^2 = 2*(-10)*d$

$d = 24.2 m$

so the total distance that it require to stop is given as

$d = 24.2 + 11 = 35.2 m$

while the deer is standing at distance 38 m

so the car will stop 2.8 m before the position of deer