Question

A ball starts from rest and accelerates at 0.465 m/s2 while moving down an inclined plane 8.20 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 16.7 m, it comes to rest. What is the speed of the ball at the bottom of the first plane?

Answer 1

Answer:

2.76m/s

Explanation:

The position x for a given time t and constant acceleration a is:

(1) $x=\frac{1}{2}at^2$

The velocity v:

(2) $v=at$

Solving equation 2 for time t:

(3) $t=\frac{v}{a}$

Combining equations 1 and 3:

(4) $x=\frac{v^2}{2a}$

Solving equation 4 for velocity v:

(5) $v=\sqrt{2ax}$

For a = 0.465m/s² and x = 8.2m:

v = 2.76m/s