Accelerating object problem

  • Thread starter rdn98
  • Start date
  • #1
rdn98
39
0
Here is the problem from one of my online homework site. :-(

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?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
970
Although your answer to (a) is "correct", I would prefer the exact answer "4/17" to "0.235".

The "clean" way to do (c) is to note that, since the initial position was at 0, x=(v0)t+(.5)(a)t^2 (as you have) and then determine when x= 0 by setting (v0)t+(.5)(a)t^2= 0 and solving for t. That's a quadratic and has two solutions. Since it factors as
t( v0+ (.5)at)= 0, one obvious solutions is t= 0 (it was at 0 when t=0) and the other is t= -v0/(.5a): the second time it is at 0.

e) shouldn't be hard. You know the general equation (you used it before) x- x0= (v0)t+ (.5)(a)t^2. In the previous problems, you had an object that started at x0= 0 with v0= 4 and a= -17 so
x= 4t- (17/2)t^2. The new object starts with x0= 0, v0= 0 and a= 0 (because it has CONSTANT speed) so for it x= 4t. You are asked when the two "paths cross". Since they are moving in a straight line, I assume that means "are at the same place at the same time". Do exactly what you said: set the equations for x equal to one another.
Since the two objects are to have the same x for the same t,
x= 4t- (17/2) t^2= 4t.
That's what you did, right? And you canceled the 4t terms to get
-(17/2) t^2= 0 which has only the solution t=0, right?

Okay- have the courage of your convictions and answer "the two objects are NEVER in the same place at the same time again!"

Think of two cars driving east at the same speed. One continues east at constant speed, the other slows steadily, comes to a stop, and then goes in reverse, going west. It gets left far behind by the first car.
 
  • #3
Pulsar
1
0
I think e) is referred to d) part in which the car has v0=8 m/s.
The answer is obviously 32/17 m.
 

Suggested for: Accelerating object problem

Replies
8
Views
553
  • Last Post
Replies
12
Views
86
  • Last Post
Replies
16
Views
708
Replies
1
Views
355
Replies
30
Views
536
Replies
7
Views
373
Top