1. No, not the average.
vf=vi+at
0=22-1.5t
t=15seconds.
Your method rounded distance to 160m. If you had not rounded, you would have gotten 15 seconds....
2.find the time of the first to get to that speed.
vf=gt solve for t, and the distance fell.
Now, subtract one from that t, and find out how far the second fell.
Subtract the distances.
1) A car traveling at 22m/s decelerates at a constant 1.5m/s^2. Calculate the time it takes to stop.
I first calculated the distance which is 160 m. Then I used the formula d=vit+1/2at^2. I got a quadratic -0.75t^2+22t-160. I got 13 or 16 as my time, but the answer says 15. Would it be correct to take the average of those times?
2) A stone is dropped from the roof of a high building. A second stone is dropped 1s later. How far apart are the stones when the second one has reached a speed of 23.0m/s?
I set this problem up by writing the given for stone 1 and stone 2.
Some equations I got were d(s1)= 4.905t^2 and ts2=ts1+1
I don't know how to do the problem.
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