The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.

Question

i. Compute the acceleration.
ii. How much further will the Bus travel before coming to rest, provided the acceleration remain constant?

Answer 1

15t + 1/2 at^2 = 90
Now, a = ∆v/∆t = -8/t
15t - 4t = 90
11t = 90
t = 90/11 s
a = -8/(90/11) = 0.98 m/s^2

so, it will take another 7/0.98 = 7.16 seconds to stop, covering a distance of
7*7.16 - 0.49 * 7.16^2 = 25 meters

Answer 2

can u explain what did u do at the last please? and where did that 0.49 come from?

Answer 3

d=vf*t-1/2 a t^2=7*7.16 - 0.49 * 7.16^2 = 25 meters