Question

A zebra starts from rest and accelerates at 1.9 m/s2. How far has the zebra gone after 5 seconds?

Answer 1

First, find the final speed: Vf = Vi + at
Vi = 0 m/s
a = 1.9 m/s^2
t = 5
Vf = 0 + (1.9)(5) = 9.5 m/s

Second, find the distance travelled: Vf^2 = Vi^2 + 2a(Xf - Xi)
Vf = 9.5 m/s
Vi = 0 m/s
a = 1.9 m/s^2
Xi = 0 m

9.5^2 = 0 + (2)(1.9)(Xf - 0)
90.25 = 3.8 Xf
Xf = 23.75 m

The zebra goes 23.75 m.

Answer 2

s(t) = 1/2 a t^2 + vo t + so

since velocity at rest is 0, and the starting position is 0

s(t) = 1/2 a t^2, for some constant acceleration, a