Stuck on a Mechanics 2 question!

2014_GCSE

"A particle moves in a horizontal plane and its position vector at time t is relative to a fixed origin O is given by:

r = (2 sin t)i + (cos t)j

Find the values of t in the range "0 <= t <= pi" when the speed of the particle is a maximum.

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Stuck on this... any help and a walkthrough on what to do?

Reply 1

troubadour.

Original post by 2014_GCSE

We're not allowed to post full solutions but I'll say this: remember that differentiating the position vector gives a velocity vector and differentiating a velocity vector gives an acceleration vector. Also remember that acceleration is zero when the particle is at its maximum speed.

Reply 2

2014_GCSE

Original post by Hydeman

We're not allowed to post full solutions but I'll say this: remember that differentiating the position vector gives a velocity vector and differentiating a velocity vector gives an acceleration vector. Also remember that that acceleration is zero when the particle is at its maximum speed.

Haha, this is as far as I got!
But okay, that rule seems fair.

Here's what I have so far:

So I differentiated twice and got

a = (-2 sin t)i + (-cos t)j

And I'm not quite sure how to set this equal to 0 exactly.
Do I find the acceleration magnitude and put THAT equal to 0?

Reply 3

troubadour.

Original post by 2014_GCSE

Huh. I'm stuck on that too. xD Well, technically the magnitude of the acceleration is zero at the time that you want to work out but I don't understand how you could get further than:

a = [(-2 sin t)² + (-cos t)²]^0.5 = 0

Reply 4

2014_GCSE

Original post by Hydeman

Expand to: ROOT: 4 sin^2 (t) + cos^2 (t)

Then use a trig. identity to get ROOT: 3sin^2(t) + 1= 0

Then you have sin^2(t) = -1/3

Which has no solutions... :s-smilie:

Reply 5

troubadour.

Original post by 2014_GCSE

Expand to: ROOT: 4 sin^2 (t) + cos^2 (t)

Then use a trig. identity to get ROOT: 3sin^2(t) + 1= 0

Then you have sin^2(t) = -1/3

Which has no solutions... :s-smilie:

Is this in a book? If so, do you have answers in the back?

Reply 6

2014_GCSE

Original post by Hydeman

Is this in a book? If so, do you have answers in the back?

t = 0

t = pi

Reply 7

troubadour.

Original post by 2014_GCSE

t = 0

t = pi

Huh. That's very strange... :/

I'll tag a maths teacher I know of and see if she can help you. Sorry I couldn't be of much help.

Reply 8

Muttley79

Original post by 2014_GCSE

Hi - I've just been tagged - I've noted it says 'speed' not velocity.

I found v^2 and use that trig identity to see when it would be max - try that.

Reply 9

2014_GCSE

Original post by Muttley79

Hi - I've just been tagged - I've noted it says 'speed' not velocity.

I found v^2 and use that trig identity to see when it would be max - try that.

I'm not sure why this is? But I did it and I got:

ROOT: 3 cos^2(t) + 1 = v

or
3 cos^2(t) + 1 = v^2

Reply 10

Muttley79

Original post by 2014_GCSE

I'm not sure why this is? But I did it and I got:

ROOT: 3 cos^2(t) + 1 = v

or
3 cos^2(t) + 1 = v^2

When is cos x max? You are nearly there

Reply 11

2014_GCSE

Original post by Muttley79

When is cos x max? You are nearly there

When cos x = 1

So you want x to be 0 or 2pi to get max velocity!

But the answer is t = 0 and t = pi :frown:

Reply 12

Muttley79

Original post by 2014_GCSE

When cos x = 1

So you want x to be 0 or 2pi to get max velocity!

But the answer is t = 0 and t = pi :frown:

The range only goes up to pi in the question.

Reply 13

2014_GCSE

Original post by Muttley79

The range only goes up to pi in the question.

Yeah but if the answers I got were t = 0 and t = 2pi

And the range is 0 to pi, then surely the only answer should be t = 0? Where does the t = pi answer come from?

Reply 14

Muttley79

Original post by 2014_GCSE

Yeah but if the answers I got were t = 0 and t = 2pi

And the range is 0 to pi, then surely the only answer should be t = 0? Where does the t = pi answer come from?

Can you double check the question is copied correctly? I'm busy prepping lessons for tomorrow at the moment but will come back later and check this thread.

Reply 15

2014_GCSE

Original post by Muttley79

Can you double check the question is copied correctly? I'm busy prepping lessons for tomorrow at the moment but will come back later and check this thread.