Question

A zebra starts from rest and accelerates at 1.9 m/s, how far has the zebra gone after 5 seconds

Answer 1

Answer:

23.8 m

Explanation:

The distance travelled by the zebra can be calculated by using the equation:

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

t is the time

a is the acceleration

For this zebra,

u = 0 since it starts from rest

$a=1.9 m/s^2$ is the acceleration

Substituting t = 5 s, we find the distance travelled by the zebra:

$d=0+\frac{1}{2}(1.9)(5)^2=23.8 m$