2. A sports car claims to increase its speed from rest to 25 m/s in 3.6 seconds. How far will the car go in this time?

Explanation:

we first find acceleration by

V=u +at (v is the final velocity,u the initial velocity,a the acceleration and t the time taken)

v=25 m/s

u=0

t=3.6s

substituting we get

25=0 + a (3.6)

25=3.6a (diving both sides by 3.6 we get)

a=6.94 m/s^2

to find distance we use the equation v^2=u^2 + 2ad

25^2 =0^2 + 2 (6.94)do

625=13.88d (dividing both sides by 13.88 we get );

d=45m

0.55x745.7j and a calculator