Difficult maths question/ velocity acceleration

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Your answers look fine. For the last question, it seems to me that it would be referring to the the moment when $a(t)=0$ 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 😊

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.

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 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.

"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

"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.

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.