course phy 201
An automobile traveling a straight line is at point A at clock time t = 9 sec, where it is traveling at 10 m/s, to a certain point B. If the automobile accelerates uniformly at a rate of .9 m/s/s, then if it reaches point B at clock time t = 14 sec, what is its velocity at that point? What is its average velocity over the interval from A to B? If at point A the automobile is 44 meters from the starting point, how far is point B from the starting point?Solution:
1. To find the velocity of the point when we have the uniform acceleration and time then we will use the following acceleration equation:
Acceleration = change in velocity / change in time
By rearrange the above equation
Change in velocity = Acceleration*change in time
=0.9m/s/s*14s = 12.6m/s
2. Average velocity = change in position / time
As change in position = s = v* t = 10*9 = 90
So ,
= 90m/9sec= 10m/s
3. Initial distance is = A = 44
Final distance = B= ?
Average velocity = final distance – initial distance/ time
Final distance= velocity *time + initial distance
=10m/s*9sec +44m =134 m
&#This looks very good. Let me know if you have any questions. &#