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)
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.
2 answers
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.
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.