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AS Maths - Mechanics - Constant acceleration
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A particle P is moving on the x-axis with constant deceleration 4 ms^-2.
At time t = 0, P passes through the origin O with velocity 14 ms^-1 in the positive direction.
The point A lies on the axis and OA = 22.5m. Find:
a) the difference between the times when P passes through A.
b) the total distance travelled by P during interval between these times.
(a)
s = 22.5, u = 14, v = ?, a = -4, t = ?
I solved this (s = ut + 1/2at^2) and got two answers for time.
t = 2.5 or t = 4.5
The difference (4.5 - 2.5) being 2 seconds.
(b)
I'm not sure how to answer this section.
I'm thought the variables would be: s = ?, u = ?, v = 0, a =-4, t = 2
A particle P is moving on the x-axis with constant deceleration 4 ms^-2.
At time t = 0, P passes through the origin O with velocity 14 ms^-1 in the positive direction.
The point A lies on the axis and OA = 22.5m. Find:
a) the difference between the times when P passes through A.
b) the total distance travelled by P during interval between these times.
(a)
s = 22.5, u = 14, v = ?, a = -4, t = ?
I solved this (s = ut + 1/2at^2) and got two answers for time.
t = 2.5 or t = 4.5
The difference (4.5 - 2.5) being 2 seconds.
(b)
I'm not sure how to answer this section.
I'm thought the variables would be: s = ?, u = ?, v = 0, a =-4, t = 2
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#2
(Original post by throwaway21458)
Question
A particle P is moving on the x-axis with constant deceleration 4 ms^-2.
At time t = 0, P passes through the origin O with velocity 14 ms^-1 in the positive direction.
The point A lies on the axis and OA = 22.5m. Find:
a) the difference between the times when P passes through A.
b) the total distance travelled by P during interval between these times.
(a)
s = 22.5, u = 14, v = ?, a = -4, t = ?
I solved this (s = ut + 1/2at^2) and got two answers for time.
t = 2.5 or t = 4.5
The difference (4.5 - 2.5) being 2 seconds.
(b)
I'm not sure how to answer this section.
I'm thought the variables would be: s = ?, u = ?, v = 0, a =-4, t = 2
Question
A particle P is moving on the x-axis with constant deceleration 4 ms^-2.
At time t = 0, P passes through the origin O with velocity 14 ms^-1 in the positive direction.
The point A lies on the axis and OA = 22.5m. Find:
a) the difference between the times when P passes through A.
b) the total distance travelled by P during interval between these times.
(a)
s = 22.5, u = 14, v = ?, a = -4, t = ?
I solved this (s = ut + 1/2at^2) and got two answers for time.
t = 2.5 or t = 4.5
The difference (4.5 - 2.5) being 2 seconds.
(b)
I'm not sure how to answer this section.
I'm thought the variables would be: s = ?, u = ?, v = 0, a =-4, t = 2
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(Original post by Pangol)
As you have t = 2, you are presumably considering the motion from the first time that P passes through A to the second time that P passes through A. Do you have any reason to think that the velocity of P is zero at the end of this motion? It might be better to consider the total displacement for this part of the motion.
As you have t = 2, you are presumably considering the motion from the first time that P passes through A to the second time that P passes through A. Do you have any reason to think that the velocity of P is zero at the end of this motion? It might be better to consider the total displacement for this part of the motion.
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#4
(Original post by throwaway21458)
I'm still not sure what my variables would be after reading that. Would they be: s = -22.5, u = 0, v = ?, t = 2?
I'm still not sure what my variables would be after reading that. Would they be: s = -22.5, u = 0, v = ?, t = 2?
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(Original post by Pangol)
If you're using t = 2, then you are considering the motion from A and back to A again. Would s = -22.5? No - this is the displacement from A to O, and you are not finishing at O. You are finishing at A. What is the displacement from A to A? Similarly, I don't see how you can use an initial velocity of zero. The velocity of P is not zero either of the times when it passes through A. Best not to use either of the velocities.
If you're using t = 2, then you are considering the motion from A and back to A again. Would s = -22.5? No - this is the displacement from A to O, and you are not finishing at O. You are finishing at A. What is the displacement from A to A? Similarly, I don't see how you can use an initial velocity of zero. The velocity of P is not zero either of the times when it passes through A. Best not to use either of the velocities.
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#6
(Original post by throwaway21458)
Could you just walk me through how to answer part b?
Could you just walk me through how to answer part b?
Your first idea - s = ?, u = ?, v = 0, a = -4, t = 2 - is really close. Although you don't say so, I assume you are considering the motion from A to the point where P stops moving. If you can find the s for this problem, then you just have to double it to answer the question. The only bit of fine-tuning that you need is that t is not 2 - that is the number of seconds required for P to g from A and back to A again. If P goes from A to the point where it stops, what value of et do you think you need?
Apologies for the earlier nonsense!
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