Question

You're driving down the highway late one night at 20m/s when a deer steps onto the road 54 m in front of you. Your reaction time before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s2.

Part A How much distance is between you and the deer when you come to a stop? Express your answer to two significant figures and include the appropriate units.

s =

Part B What is the maximum speed you could have and still not hit the deer? Express your answer to two significant figures and include the appropriate units.

v=

Answer 1

(A) as your reaction time is 0.50 sec, which means that you have advanced by 0.50*20 i.e. 10 m by the time of your spottting the deer and before you actually press the brakes.

That makes the actual distance between your car and the deer = (54 - 10)= 44 m at the point of your hitting the brakes.

now, applying third equation of motion,

V² - U² = 2*a*d

where V and U are the final and initial velocities of the car respectively. and a is the acceleration. and d is the distance covered during the deceleration.

given :- V = 0 m/s ; U = 20 m/s ; a = -10 m/s²

now putting the values,

0 - 20² = 2 * (-10) * d

d = 400/20

d = 20 m

so, the distance between you and the deer when you come to stop is

s = 44 - d

s = 44 - 20 = 24 m

thus the distance between you and the deer when it come to stop is 24 meters.

(B) now for the maximum speed you can cover the distance to 44 m.

now applying the same third equation of motion,

V² - U² = 2*a*S

given :- V = 0 m/s ; a = -10 m/s² ; S = 44 m

0 - U² = 2 * (-10) * 44

U² = 880

U = 29.66 m/s

thus, the maximum speed you could have and still not hit the deer is 29.66 m/s.

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