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:

Explanation:

Given

Motorcyclist speed=12 m/s  

maximum acceleration$=-6 m/s^2$

distance=39 m

Let x be the distance traveled by motorist in his reaction time

therefore remaining 39-x will be traveled with $-6m/s^2$ acceleration

$v^2-u^2=2as$

s=39-x

v=0

u=12 m/s

$0-12^2=2\left ( -6\right )\left ( 39-x\right )$

x=27 m

Therefore he traveled 27 m in his reaction time

$27=12\times t$

t=2.25 s

(b)If his reaction time is 2.56 sec

then distance traveled in his reaction time

$x_0=12\times 2.56=30.72 m$

Remaining distance 39-30.72=8.28 m

therefore its velocity when it reaches the deer

$v^2-u^2=2as$

$v^2=12^2+2\times \left ( -6\right )\times 8.28=44.64$

v=6.681 m/s