At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1 foot between t=0 seconds and t = 1 second. 1) How far does it move between t = 1 second and t = 2 seconds?

Answer: 3 ft

Explanation:

Given:

- the ball is initially at rest before it start moving at constant acceleration.

a = constant

u = initial speed = 0

Using the formula for displacement.

d = ut + 0.5at^2 ... 1

Since u = 0, equation 1 becomes;

d = 0.5at^2 ... 3

making a the subject of formula

a = 2d/t^2 ... 2

Where; a = acceleration, d = distance travelled, t = time taken, u = initial speed

d = 1 ft, t = 1 s

Substituting into equation 2 above.

a = (2*1) / 1^2 = 2ft/s^2

Since the acceleration is 2ft/s^2, at t = 2 sec the distance it would have covered will be:

Using equation 3;

d = 0.5 (2 * 2^2) = 0.5 (8) = 4ft

therefore, the distance travelled between t = 1 to t = 2

Is:

d2 = d - d1

Where d = total distance (t=0 to t = 2) = 4 ft

d1 = distance covered between (t=0 to t=1) = 1 ft

d2 = 4 - 1 = 3ft