Question

A) A motorist traveling at 12 m/s encounters a deer in the road 41 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? B) If his or her reaction time is 1.1333s, how fast will (s) he be traveling when(s) he reaches the deer?

Answer 1

Answer:

A)  Reaction Time = 2.73 seconds.

B) He will not reach the deer.

Explanation:

A) Initial speed, u = 12 m/s

   Acceleration, a = -7 m/s²

   Final velocity, v = 0 m/s

   We have equation of motion v² = u² + 2as

                0² = 12² + 2 x -7 x s

                s = 10.29 m

  He need 10.29 m to stop.

  Remaining distance = 43 - 10.29 = 32.71 m

  We have

            Remaining distance = Reaction Time x Initial velocity

            32.71 = Reaction Time x 12

            Reaction Time = 2.73 seconds.

B) Distance traveled in 1.1333 s = 1.1333 x 12 = 13.60 m

   Remaining distance = 41 - 13.6 = 27.4 m

    Initial speed, u = 12 m/s

   Acceleration, a = -7 m/s²

   Displacement, s = 27.4 m

                      v² = 12² + 2 x -7 x 27.4

                      v² =  -239.6

   Not possible.

   Motorist will not hit deer.

  He will not reach the deer.