Ball released from rest at the top of slope

[electrofanner]

Homework Statement

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.

 

[Doc Al]

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.

The speed uniformly increases from 0 to some final speed V_f. So the average speed is:
V_ave = (0 + V_f)/2 = V_f/2.

(If that's not convincing enough, you can also derive this fact more formally using standard kinematics equations.)

 

[electrofanner]

Thanks :)