Kinematics TPR HL #39

[Addallat]

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Question:

A particle moving along the x-axis passes through the point x = 0 (in the -x direction) at a particular instant. If it experiences a constant acceleration of -2 m/s^2, where could the object be 3 seconds later?


A. x = -3 m
B. x - 6 m
C. x = -9 m
D. x= -12 m


I couldn't follow the explanation in the book. Can any of you guys post your work for solving this problem. Do you have to solve for Vo first?

I could see that solving for Vo would be X = Vo(t)+1/2(a) t^2
I know X = 0
a = -2 m/s^2
but what do you use for time? The book states that you should use t= 3s to solve for Vo, which really confuses me...why would t = 3 seconds when x = 0? I thought the entire purpose of the question is to solve for displacement when time = 3 seconds


If I had Vo then I would just plug in to solve for x = Vo(t)+1/2(a)t^2

x = ?(3s) + 1/2 (-2 m/s^2) (3s)^2

Answer is D

 

[Addallat]

I'm still stumped by this question. Am I lacking some concept here? Why can you assume that time = 3 seconds when the object is at the zero mark? I really can't see how you go about getting the initial velocity at the zero mark

 

[onedirection]

I mean it was traveling in the - direction, therefore velocity is already a - value. The - acceleration would increase the negativeness of the velocity

If initial velocity were 0

Then it would have ended up at -9

But since it already had an initial velocity, you know that it must have gone farther than -9 because the final velocity would be greater than it would have been than if it were at a 0 initial velocity

The only answer choice where the magnitude of x is > 9 is choice D

I kind of wasted time trying to do math at first as well, then you realize that the answer is pretty obvious when you think about it. Taught me a lesson there too