Amotorist 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.

Explanation:

Given

Motorcyclist speed=12 m/s  

maximum acceleration

distance=39 m

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

therefore remaining 39-x will be traveled with acceleration

s=39-x

v=0

u=12 m/s

x=27 m

Therefore he traveled 27 m in his reaction time

t=2.25 s

(b)If his reaction time is 2.56 sec

then distance traveled in his reaction time

Remaining distance 39-30.72=8.28 m

therefore its velocity when it reaches the deer

v=6.681 m/s

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

see earlier answer