A zebra starts from rest and accelerates at 1.9 m/s2 what will be the acceleration after 5 seconds

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

Corrected question;

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

Explanation:

From the equation of motion;

d = vt + 0.5at^2 1

Where;

d = displacement

v = initial velocity = 0

t = time taken= 5 seconds

a = acceleration = 1.9 m/s^2

Substituting the given values into equation 1;

d = 0(5) + 0.5(1.9×5^2)

d = 0.5(1.9×5^2)

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

answer; (arch); good luck

i guess it’s a and b