course Phy231
2/19/10 7pm
At clock time t = 7 sec, a ball rolling straight down a hill is moving at 6 m/s and is 47 m from the top of the hill. It accelerates uniformly at a rate of .6 m/s/s until clock time t = 17 sec.
What is its velocity at this point and what is its average velocity during this time?
How far is it from the starting point at t = 17 sec?
17s * .6m/s/s= 10.2m/s So at 17sec the velocity is 10.2m/s +6m/s= 16.2m/s
The time interval lasts only 10 seconds, from t = 7 sec to t = 17 sec. You don't have any information about the motion before or after this interval.
vAve= vf - v0= 16.2m/s + 6m/s= 22.2m/s 22.2m/s /2= 11.1 m/s
Had the 16.2 m/s been correct, you would have obtained the correct average velocity.
7sec....47m
8sec....53.6m
9sec....60.8m
10sec...68.6m
11sec...77.0m
12sec...88.0m
13sec...97.6m
14sec...107.8m
15sec...118.6m
16sec...130.0m
17sec...142.0m from the starting point"
This is very close to what would happen on each 1-second interval, though your displacements are based on the final velocity rather than the average velocity for each so you've overestimated by a small amount. It is often necessary to do an interval-by-interval approximation of this type, but it isn't necessary when acceleration is uniform, and in this case it simply isn't practical. You need a more general approach.
If you multiply the correct average velocity (obtained using the 10 second time interval) you will get the correct displacement, which I believe will be 90 m, and will make your final position 90 m + 47 m = 137 m rather than your 142 m.