Question

A motorist traveling at 12 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −6 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s. 015 (part 2 of 2) 10.0 points If his or her reaction time is 2.56 s, how fast will (s) he be traveling when (s)he reaches the deer? Answer in units of m/s

Answer 1

Answer:

2.25 seconds

6.68 m/s

Explanation:

t = Time taken

u = Initial velocity = 12 m/s

v = Final velocity

s = Displacement = 39 m

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-6}\\\Rightarrow t=2\ s$

The car will stop in 2 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 2+\frac{1}{2}\times -6\times 2^2\\\Rightarrow s=12\ m$

The distance the car will cover is 12 m

Distance between the car and deer will be 39-12 = 27 m

Time = Distance / Speed

Maximum reaction time will be 27/12 = 2.25 seconds

If her reaction time is 2.56 seconds

The distance she will travel in that time

Distance = Speed × Time

⇒Distance = 12×2.56 = 30.72 m

Distance to the deer = 39-30.72 = 8.28 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -6\times 8.28+12^2}\\\Rightarrow v=6.68\ m/s$

The car will be traveling at 6.68 m/s