Question

A car accelerates uniformly from rest and after 12 seconds has covered 40m.

What are its acceleration and its final velocity?

Answer 1

`a\approx 0.56 \,(m)/(s^2)`

`v_f=6.72\,(m)/(s)`

Step-by-step explanation:

This is a motion under constant acceleration, so we can use the kinematic equation for the distance covered:

`x_f-x_i=v_i\,t+(1)/(2) a\,t^2\\40=0+(1)/(2) a\,(12)^2\\40=72 \,a\\a=(40)/(72) \,(m)/(s^2) \\a\approx 0.56 \,(m)/(s^2)`

Now, the final velocity can be calculated via:

`v_f=v_i+a\,t\\v_f=0+0.56\,(12)\\v_f=6.72\,(m)/(s)`