Question

How far will a car travel while it accelerates for 3.2 seconds at a rate of 2.5m/s^2 if it's velocity is 9.8 m/s?

Answer 1

Answer:

Car travel distance

s(t)=(at^2)/2+v0t,

the speed is

v(t)=v0+at,

where a is acceleration, v0 is initial speed.

Here

a=2.5m/s^2

t=3.2s

v(t=3.2s)=9.8m/s.

From hence

v0=v(t)-at=9.8m/s-8m/s=1.8m/s

s=18.56m