first do the stopping problem
d = Vi t + .5 a t^2
d = 25 t + .5 a t^2
50 = 25 t + .5 a t^2
v = Vi + a t
0 = 25 + at
a t = -25
a = -25/t
combine 50 = 25 t - .5(25t)
50 = .5 (25 t)
t = 4 seconds to stop after applying brakes
a = -25/4 while stopping
Now do the problem
for the first part
x = 25 t until t = 2 where x = 50
then x = 50 + 25(t-2) - (25/8) (t-2)^2
until stopped at x = 100 and t = 6
during deacceleration v = 25 -(25/4)(t-2)
a stunt car driver testing the use of air bags drives a car at a constant speed of 25m/s for a total of 100 meters. he applies his brakes and accelerates uniformlu to a stop just as he reaches the wall 50 meters away. SKETCH THE A GRAPH SHOWING POSITION V.S TIME and VELOCITY v.s TIME
also how long does it take the car to travel the first 100 meters?
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