A car traveling at initial velocity 25 meters per second slows down and stops in

Question

A car traveling at initial velocity 25 meters per second slows down and stops in a distance of 90 meters. a) Find the acceleration of the car while stopping. b) Find the speed the car had at the 40 meter point of the trip. A car traveling at 31 meters per second (about 70 miles per hour) slams on the brakes and decelerates at a rate of 7 m/s^2. a) Find the distance the car will travel until it stops. b) If this car slammed on its brakes when it was half a football field length (50 yards) away from a deer in the road, did the car stop before hitting the deer?

Explanation / Answer

Solution:

7) .a) Initial velocity = vi = 25m/s

final velocity = vf=0 (stops).

distance traveled = d = 90 m

using the kinematic equation vf^2 =vi^2+2ad , acceleration a is calculated

a = (vf^2 -vi^2) /2d = (0^2 - 25^2) / 2*90 = - 3.5 m/s^2

b) when d=40m, vf^2 = vi^2 +2ad

vf^2 = 25^2 +2*(-3.47)(40) = 347.2

vf = 18.6 m/s = velocity at the 40 m distance.

8) a) velocity of the car = 31 m/s

deceleration = a = -7 m/s^2

final velocity = vf = 0

distance d = vf^2-vi^2/2a = (0^2-31^2)/2*(-7) = 68.6 m

b) distance to the deer =d1= 50 yards = 45.7 m

velocity of the car at the point of applying brakes = vi'

distance traveled = d1=68.6 - 45.7 = 22.9 m

vi^2 = vi^2+2ad

=31^2 +2(-7)(22.9)

=> velocity at the end of 22.9 m (point where the deer is sighted) = 25.3 m/s

vi' = 25.3 m/s

vf=0 a= -7m/s^2

d= [0^2 - (25.3)^2 ] / 2*(-7) = 45.7 m

So the car stops before hitting the deer.

  

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