a car is braked so that it slows down with uniform retardation from 15 m/s to 7 m/s while it travels a distance of 88m. if the car continues to slows down at same rate ,aftervwhat further distance will it be further be brought to rest.

Answer 1

average velocity is 11 m/s ... (15 + 7) / 2

time is 8 s ... 88 / 11

acceleration is -1 m/s^2 ... (7 - 15) / 8

time to stop is 7 s ... -7 / -1

ave stopping velocity is 3.5 m/s
... (7 + 0) / 2

stop distance = (stop time) * (ave stop vel)

Answer 2

V^2 = Vo^2 + 2a*d.
7^2 = 15^2 + 2a*88
-176a = 15^2-7^2
a = -1.0 m/s^2.

0 = 15^2 + 2*(-1)d.
d = 112.5 m.

112.5 - 88 = 24.5 m. further.