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?Ok I drew a sketch for this problem but it only confused me more so I think I can just plug in the equations to get these answers.
1. What is its velocity at that point? I used the equation v= v0 + at so by filling in the equation v = 10m/s + (.9m/s/s)(14s-9s) so v = 10 m/s + (.9m/s)(5s) so v = 10m/s + 4.5 m/s v = 14.5m/s
2. What is its average velocity over the interval from a to b? vAve = (vf+v0)/2 so vAve = (14.5 m/s +10m/s)/2 vAve = 12.25m/s
3. How far is point b from the starting point? I think you use the equation x=x0 +v0t+1/2at^2 so x = 44m +(10m/s)(9sec) + ?.9m/s/s)(14s)^2 so x = 44m + 90m/s/s +88.2m so x = 222.2 meters.
The equation is good, but it only holds if the acceleration is uniform for the entire interval. We don't know if the acceleration was uniform prior to the t = 9 sec instant, so we can't apply the equation here (at least not easily).
All the information we have is for the 5-second interval between t = 9 s and t = 14 s. For that interval we know that `dt = 5 sec, v0 = 10 m/s and vf = 14.5 m/s. Thus vAve = (10 m/s + 14.5 m/s) / 2 = 12.75 m/s, and `ds = vAve * `dt = 12.75 m/s * 5 s = 63.75 m.
you did good work here, but be sure you understand my note