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In an experiment, a ball was released from rest at the top of a slope and rolled a distance of 2.0 m down the slope in 8.0 s. Calculate

(a) its average speed in this time

(b) the speed after 8.0s from rest

(c) the acceleration of the ball

Basically, I got the answer for a...

Which is simply Distance_{Tot.}divided by Time Taken

= 2.0 m/8.0 s

= 1/4 m/s

The answer for b is 0.50 m/s. I have come to a conclusion that the average speed is x0.5 of the speed at that particular instance.

(2x Avg. speed in that 8s = Speed of object in 8s)

However, I don't get why this came to be. Someone please explain to it to me. Thanks

For c... Simple also (after getting b)

Which is Velocity/Time

0.50 m/s divided by 8s = 0.0625 m/s^{2}

Basically, All I need is an explanation for Part b.

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# Ball released from rest at the top of slope

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