Kinematics problem -- A particle is moving with changing acceleration...

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Homework Statement:
please help on kinematics problem
Relevant Equations:
v=ds/dt, a=dv/dt
A particle, P, starts from rest at a point X and moves in a straight line with an acceleration expressed as a=4t. After 2 seconds, the particle reaches Y and it stops accelerating. The particle leaves Y with a velocity -3ms-1, and finally comes to rest at Z.
(i) Find the value of t when the particle reaches Z. (answer is t=3.06s)
 

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  • #2
PeroK
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Problem Statement: please help on kinematics problem
Relevant Equations: v=ds/dt, a=dv/dt

A particle, P, starts from rest at a point X and moves in a straight line with an acceleration expressed as a=4t. After 2 seconds, the particle reaches Y and it stops accelerating. The particle leaves Y with a velocity -3ms-1, and finally comes to rest at Z.
(i) Find the value of t when the particle reaches Z. (answer is t=3.06s)

That problem doesn't make sense to me. Are you sure you've written it correctly?

As with your other post, you need to show us how much of this you can do.
 
  • #3
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That problem doesn't make sense to me. Are you sure you've written it correctly?

As with your other post, you need to show us how much of this you can do.
I got this question from my reference book. I cant get the answer.
 
  • #4
PeroK
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I got this question from my reference book. I cant get the answer.

This is not a forum where we do your homework for you. Our approach is to help you where you get stuck. This involves you doing as much of the problem as possible, so we can see where you are going wrong.

In this case, however, the problem is clearly nonsensical. If that's what your book really says, then I'd skip this problem.
 
  • #7
HallsofIvy
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A particle, P, starts from rest at a point X and moves in a straight line with an acceleration expressed as a=4t.
Assuming "t" here is the time, in seconds, after the particle started moving (which should have been explicitly said) then its speed after t seconds is [tex]2t^2[/tex] and it will have traveled a distance of [tex]\frac{2}{3}t^3[/tex].

After 2 seconds, the particle reaches Y and it stops accelerating.
Taking X to be 0, [tex]Y= \frac{2}{3}2^3= \frac{16}{3}[/tex] and the particle's speed at Y is 2(4)= 8.

The particle leaves Y with a velocity -3ms-1, and finally comes to rest at Z.
This is the part that makes no sense! If the particle has the constant velocity [tex]-3 m/s[/tex] then it will never come to rest!

If, instead, the particle left Y with acceleration [tex]-3 m/s^2[/tex] then its velocity t seconds after leaving Y is [tex]\frac{16}{3}- 3t[/tex] which will be 0 when [tex]t= \frac{16}{9}[/tex]. That is not "3.06" so apparently that is not a correct interpretation either!

Another possibility is that something happened at Y so that the velocity dropped instantaneously from 8 to -3 m/s while there was still the acceleration 4t. Then the speed t seconds after leaving Y is [tex]-3+ 2t^2[/tex]. The particle will stop when [tex]-3+ 2t^2= 0[/tex] so [tex]t= \sqrt{\frac{3}{2}}[/tex]. No, that's not "3.06" either! I am baffled by this problem.
 

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