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

Step-by-step 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* t`

t=2.25 s

(b)If his reaction time is 2.56 sec

then distance traveled in his reaction time

`x_0=12* 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* \left ( -6\right )* 8.28=44.64`

v=6.681 m/s