A car accelerates at a rate of 6m/s^2 for 4 seconds until it has traveled a total of 40m what was the initial velocity of the car

Using the 2nd equation of motion;

= s = ut + ½at²

= 40 = u×2 + ½ × 6 × 2²

= 40 = 2u + 3 × 4

= 40 - 12 = 2u

= 28/2 = u

= 14 m/s = u

And its done! Simple isn't? :P

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.