- #1

rdn98

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An object starts at time t = 0 with a velocity of v0 = +4 m/s and undergoes a constant acceleration of a = -17 m/s2.

There's five small parts to this problem but I got 4/5 done. I'll post them anyway to give you guys a clearer picture of what's happening here.

a) At what time does the speed of the object reach zero?

I used the equation v=vo+at. and I got the correct answer of t1=.235 seconds.Ok, that was easy.

b) How far from its starting (t = 0) position is the object at time t1?

This one I used the equation x-x0=(v0)t+(.5)(a)t^2

where x0=0, t=.235, a=-17

Answer I got was D=470 meters. Ok, that was easy too.

c) At what time does the object again pass through the starting (t = 0) position?

This I wasn't sure about. I figured if it took .235 seconds to reach a velocity of 0, then another .235 seconds would make it zero again. I typed in the answer .470 seconds. It's right, but I would like to know if this is the right reasoning, or is there another mathematical way of getting the correct answer?

d) If the object had initially been moving twice as fast (8 m/s), how far would it have gone before its velocity reached zero?

This I did the same thing with problem 1 and two, but used a different velocity number. The answer is 1.88m.

Now, this is where I get stuck...

e) Suppose a second object begins moving with a constant speed of v = 4 m/s in the same direction from the same location at the same time as the object in part (d). At what time do the paths of these two objects once again cross?

I know I have to derive equations which describe the x-position of both objects and set these equal to each other and solve for time. But I don't know how to derive the equations. I tried setting up two equations equal to each other, but the time I get is 0, which is wrong. Any ideas?