# Difficult maths question/ velocity accelerationWatch

#1
Hello, I do not fully understand how to solve this question.

The one-dimensional displacement (S meters) of a particle after t seconds is given by the function s=t(t-4)^2

1. Find the function representing the velocity of the particle at time t
2. At what time is the particle at rest?
3. At what time does the particle have zero acceleration?
4. What is the particle doing at this time and where is it?

I have tried to solve these questions and have answered some of them but would appreciate any help in explaining the answer and how to reach it. I have read my notes and several textbooks sections but I have only found a few examples in a different format which do not seem to be relevant to this particular question.

For the second question I think you would solve it by differentiating the function s=t(t-4)^2. So if the particle is at rest the velocity=0.
Differentiate the function to 3t^2-16t+16=0. Then factor to (3t-4)(t-4) to find that t=4/3 or t=4. So the particle is at rest after 4/3 seconds and again after 4 seconds.

For the third question the acceleration =0. Which is substituted into the function. And at this time the velocity would reach a maximum or minimum because a=dv/dt=0.

I am not sure on how to solve the last question or if the answers I have given for the previous are correct. However, any help in finding the answers would be highly valued
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1 year ago
#2
(Original post by JS547)
....
Your answers look fine. For the last question, it seems to me that it would be referring to the the moment when the time you found in part 3. It wants you to determine the velocity and displacement at this time.
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#3
(Original post by RDKGames)
Your answers look fine. For the last question, it seems to me that it would be referring to the the moment when the time you found in part 3. It wants you to determine the velocity and displacement at this time.
Thank you for your reply. I am not sure how to find the velocity and displacement at that time. Could you perhaps elaborate.
Thank you for the help 😊
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1 year ago
#4
(Original post by JS547)
Thank you for your reply. I am not sure how to find the velocity and displacement at that time. Could you perhaps elaborate.
Thank you for the help 😊
For the displacement, you want to substitute your value for t into the original equation s=...

For the velocity, you want to substitute your value for t into the velocity function you worked out in part 1.
#5
(Original post by ghostwalker)
For the displacement, you want to substitute your value for t into the original equation s=...

For the velocity, you want to substitute your value for t into the velocity function you worked out in part 1.
Thank you for your reply. For part 1 do you mean the question asking to find the function representing the velocity of the particle at time t? If so, I do not really understand how to do this, could you explain at all how to solve that.
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1 year ago
#6
(Original post by JS547)
Thank you for your reply. For part 1 do you mean the question asking to find the function representing the velocity of the particle at time t? If so, I do not really understand how to do this, could you explain at all how to solve that.
Yes.

There is nothing to solve.

In part 1 you found the velocity as a function of time.

You now want to find the velocity at a particular time.

So, substitute that particular time for the "t" in that function you found, and work out the value of the function.
#7
(Original post by ghostwalker)
Yes.

There is nothing to solve.

In part 1 you found the velocity as a function of time.

You now want to find the velocity at a particular time.

So, substitute that particular time for the "t" in that function you found, and work out the value of the function.
Ok, so the times I found are after 4/3 second and after 4 seconds which would then be substituted into the original function s=t(t-4)^2.

For 4/3 second :

4/3(4/3-4)^2= 256/27 or 9.5 to 1dp.

s= 9.5

For 4 seconds:
4 (4-4)^2=0
s = 0
0
1 year ago
#8
(Original post by JS547)
Ok, so the times I found are after 4/3 second and after 3 seconds which would then be substituted into the original function s=t(t-4)^2.

For 4/3 second :

4/3(4/3-4)^2= 256/27 or 9.5 to 1dp.

s= 9.5

For 3 seconds:
3 (3-4)^2 = 3
s = 3

To be clear does the s represent the velocity ?
"s" represents the displacement - the original formula given in the question.

Velocity, v=ds/dt - the formula you worked out in part 1.

And acceleration a=dv/dt - which you may have used for part 3.

The time mentioned in part 4, is refering to the time worked out in part 3, i.e. when the acceleration is zero. And that should be 8/3 by my calculations.

Note: Don't know where you time of 3 comes from, as it's not the answer to any part of the question. Part 1 should have times of 4/3 and 4.
#9
(Original post by ghostwalker)
"s" represents the displacement - the original formula given in the question.

Velocity, v=ds/dt - the formula you worked out in part 1.

And acceleration a=dv/dt - which you may have used for part 3.

The time mentioned in part 4, is refering to the time worked out in part 3, i.e. when the acceleration is zero. And that should be 8/3 by my calculations.

Note: Don't know where you time of 3 comes from, as it's not the answer to any part of the question. Part 1 should have times of 4/3 and 4.
Thank you this has helped enormously. Sorry I just mis-typed 4 as 3 and did not realise!

So after 4 seconds:

4(4-4)^2 = 0
s=0
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#10
(Original post by ghostwalker)
"s" represents the displacement - the original formula given in the question.

Velocity, v=ds/dt - the formula you worked out in part 1.

And acceleration a=dv/dt - which you may have used for part 3.

The time mentioned in part 4, is refering to the time worked out in part 3, i.e. when the acceleration is zero. And that should be 8/3 by my calculations.

Note: Don't know where you time of 3 comes from, as it's not the answer to any part of the question. Part 1 should have times of 4/3 and 4.
When you said to substitute the particular time where you referring to the times 4/3 and 4 seconds because after those times the particle was at rest, so would I need to find times when the particle is moving but the acceleration is 0?
So, if a = dv/ dt
a = 6t - 16
And when the acceleration =0
6t -16 =0
16/6 = 8/3
The acceleration is at zero when t = 8/3
0
1 year ago
#11
(Original post by JS547)
When you said to substitute the particular time where you referring to the times 4/3 and 4 seconds because after those times the particle was at rest,
The particle in only at rest AT those times. Either side of those times, it's moving.

so would I need to find times when the particle is moving but the acceleration is 0?
So, if a = dv/ dt
a = 6t - 16
And when the acceleration =0
6t -16 =0
16/6 = 8/3
The acceleration is at zero when t = 8/3
As I said before, for part 4 you want the time you worked out in part 3, when the acceleration is zero. What is its velocity at that time, and what is its displacement?
#12
(Original post by ghostwalker)
"s" represents the displacement - the original formula given in the question.

Velocity, v=ds/dt - the formula you worked out in part 1.

And acceleration a=dv/dt - which you may have used for part 3.

The time mentioned in part 4, is refering to the time worked out in part 3, i.e. when the acceleration is zero. And that should be 8/3 by my calculations.

Note: Don't know where you time of 3 comes from, as it's not the answer to any part of the question. Part 1 should have times of 4/3 and 4.
For part 4 ( which is worth 6 marks) I think I would have to find the speed of the particle at the time found in part 3, explain what is was doing and find maximum or minimum value of the speed.
Firstly, a=0 when t=8/3
Knowing this substitute 8/3 into the equation to find v:
v=ds/dt = 3t^2-16t+16
v = (3 x 8/3)^2 - (16 x 8/3) +16
v = ( 3 x 2 x 8/3) - (16x 8/3) +16
v= 16 - 128/3 +16
v =-32/3

The minus sign means that the particle is moving -32/3 units/ sec in a negative direction.

Find the displacement using the original function s=t(t-4)^2
So when t=8/3
8/3 (8/3 -4)^2
s=128/27
So the particle is displaced 128/27 meters or 4.7 m to 1dp.

As previously mentioned, since a = dv/ dt = 0 the speed reaches a maximum or mimimum value of v.
Would v=-32/3 be the minimum value of v?
0
1 year ago
#13
(Original post by JS547)
As previously mentioned, since a = dv/ dt = 0 the speed reaches a maximum or mimimum value of v.
Would v=-32/3 be the minimum value of v?
Since v is a quadratic with a positive t^2 term, it will be a minimum.
#14
(Original post by ghostwalker)
Since v is a quadratic with a positive t^2 term, it will be a minimum.
Thank you very much for all of your help 😊
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