# Find Distance with Acceleration and Time

I'm working on Physics homework and have run into a problem I can't figure out. Any help would be much appreciated!

Problem:

Allison exerts a steady net force of 50N on a 22-kg shopping cart initially at rest for 1.2s .

A) Find the acceleration of the cart.

I figured this out just fine. The answer is 2.3 m/s^2.

B) Find a distance it moves.

This is what I tried:

acceleration = change in velocity / change in time
2.3 m/s^2 = change in velocity / 1.2 s
change in velocity = 2.76 m/s

distance = [(initial velocity + final velocity) / 2] x time
distance = (2.76 m/s / 2) x 1.2 s
distance = 1.38 m/s x 1.2 s
distance = 1.656 m

Unfortunately, this is not the correct answer. Any thoughts on what I'm doing incorrectly?

## 1 Answer

force = acceleration * mass, hence acceleration will be $a=50N/22kg \approx 2.27m/s^2$

Distance it moves might be found by integration: $\int_0^{1.2}v(t)dt=\int_0^{1.2}atdt$, since speed $v(t)=at$

Answer to (B) then is 1/2*1.2*(2.27*1.2)=1.63, which seems pretty close to what you have got