Asked by claire
A car brakes at constant acceleration from a velocity 226m/s to 12.5m/s over a distance of 105m.find;
a)How much time elapses during this time interval.
b)what is the acceleration.
c)if the car continued to be braked at same constant acceleration how much longer would it take for it to stop and what additional distance would it cover?
v=u+at 12.5=226+10t =21.3s
s=ut+0.5at^2 a=20.4
a)How much time elapses during this time interval.
b)what is the acceleration.
c)if the car continued to be braked at same constant acceleration how much longer would it take for it to stop and what additional distance would it cover?
v=u+at 12.5=226+10t =21.3s
s=ut+0.5at^2 a=20.4
Answers
Answered by
Damon
v = Vi + a t
12.5 = 226 + a t
so
a = - 213.5/t
d = Vi t + .5 a t^2
105 = 226 t - .5 * 213/t * t^2
105 = 226 t - 106.5 t
t = .879 seconds
a = -213/.879 = -242 m/s^2
you can do the rest but suspect you have a typo
12.5 = 226 + a t
so
a = - 213.5/t
d = Vi t + .5 a t^2
105 = 226 t - .5 * 213/t * t^2
105 = 226 t - 106.5 t
t = .879 seconds
a = -213/.879 = -242 m/s^2
you can do the rest but suspect you have a typo