Question

A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second later, the ball reaches the bottom of the incline.
(a) What was the balls acceleration?(assume that the acceleration was constant).
(b) How long was the incline?

Answer 1

Answer:

(a) a = 2.44 m/s²

(b) s = 63.24 m

Explanation:

(a)

We will use the second equation of motion here:

$s = v_it+\frac{1}{2}at^2$

where,

s = distance covered = 47 m

vi = initial speed = 0 m/s

t = time taken = 6.2 s

a = acceleration = ?

Therefore,

$47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}$

a = 2.44 m/s²

(b)

Now, we will again use the second equation of motion for the complete length of the inclined plane:

$s = v_it+\frac{1}{2}at^2$

where,

s = distance covered = ?

vi = initial speed = 0 m/s

t = time taken = 7.2 s

a = acceleration = 2.44 m/s²

Therefore,

$s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2$

s = 63.24 m