M2 Maximum Vel. via Differentiation Watch

Dggj_19
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If needed, the answer for a) is t = 2, t = 5.

Concerning part b), when velocity is a minimum, dv/dt will be 0, right? Thus, the acceleration will be zero. This much I understand. But how do you find the maximum value of velocity via differentiation? Thanks for any input
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Notnek
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(Original post by Dggj_19)
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If needed, the answer for a) is t = 2, t = 5.

When velocity is a minimum, dv/dt will be 0, right? Thus, the acceleration will be zero. This much I understand. But how do you find the maximum value of velocity via differentiation? Thanks for any input
Try drawing the graph of v against t and it should become clear.
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Dggj_19
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(Original post by Notnek)
Try drawing the graph of v against t and it should become clear.
My reasoning is that for an interval of 0 =< t =< 4, it's greatest value of y (v) graphically is just the intercept, 20, which is correct. The working out in the mark scheme, however, I don't understand. It sets the acceleration equal to 0 to find a value of t for which a maximum occurs, and this is the reason why I was confused, because I was under the impression that it was only for minimum velocity that dv/dt can be 0, right? The gradient of x = 0 on this specific graph doesn't have a gradient (acceleration) of 0, either. I'm more looking for a general solution, when you have different ranges. Would questions that have different ranges / functions of t require you to graph them out and judge for yourself?
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Notnek
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(Original post by Dggj_19)
My reasoning is that for an interval of 0 =< t =< 4, it's greatest value of y (v) graphically is just the intercept, 20, which is correct. The working out in the mark scheme, however, I don't understand. It sets the acceleration equal to 0 to find a value of t for which a maximum occurs, and this is the reason why I was confused, because I was under the impression that it was only for minimum velocity that dv/dt can be 0, right? The gradient of x = 0 on this specific graph doesn't have a gradient (acceleration) of 0, either. I'm more looking for a general solution, when you have different ranges. Would questions that have different ranges / functions of t require you to graph them out and judge for yourself?
Can you please post the mark scheme?

It depends on the function/graph which method you use. E.g. if it was a negative quadratic then the maximum would be the stationary point (if it's within the domain of t) and the minimum would be one of the two end-points (you may need to check both). If it was a cubic then the maximum could be a stationary point or an end point, you'd need to check them all.

For these questions you should always do a rough sketch of the graph first to understand what's going on.
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Dggj_19
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(Original post by Notnek)
Can you please post the mark scheme?

It depends on the function/graph which method you use. E.g. if it was a negative quadratic then the maximum would be the stationary point (if it's within the domain of t) and the minimum would be one of the two end-points (you may need to check both). If it was a cubic then the maximum could be a stationary point or an end point, you'd need to check them all.

For these questions you should always do a rough sketch of the graph first to understand what's going on.
Sure, here's the excerpt:
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Every method specifically talks about evaluating the minimum rather than the maximum.
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Notnek
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(Original post by Dggj_19)
Sure, here's the excerpt:
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Every method specifically talks about evaluating the minimum rather than the maximum.
Ah I just re-read the question and realised it's asking for the greatest speed not the greatest velocity. So you'd have to check the highest and lowest points of the graph to see which one has the greatest magnitude. Make sense?
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Dggj_19
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(Original post by Notnek)
Ah I just re-read the question and realised it's asking for the greatest speed not the greatest velocity. So you'd have to check the highest and lowest points of the graph to see which one has the greatest magnitude. Make sense?
Not immediately, no. - velocity is simply a vector of a speed with direction. Direction is not the focus of this question, so why would you need to make a disticntion between velocity and speed? The logical reasoning for the correct answer still seems to me to be 'the point between 0 and 4 with the greatest velocity'. The question states that the particle only moves along the X axis positively, for any positive value of t. Does this not remove the need to account for direction in the question?
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(Original post by Dggj_19)
Not immediately, no. - velocity is simply a vector of a speed with direction. Direction is not the focus of this question, so why would you need to make a disticntion between velocity and speed? The logical reasoning for the correct answer still seems to me to be 'the point between 0 and 4 with the greatest velocity'. The question states that the particle only moves along the X axis positively, for any positive value of t. Does this not remove the need to account for direction in the question?
E.g. let's say the greatest velocity of the particle was 4 and the smallest velocity was -5. So you can imagine 4 at the top of the graph and -5 at the bottom.

A velocity of -5 gives you a speed of +5 (speed is always positive i.e. the magnitude of veliocity) which means that in this case the "greatest speed" of the particle would be 5 but the "greatest velocity" would be 4.

EDIT: "Greatest" velocity could be a confusing term to use depending on the question.
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Dggj_19
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E.g. let's say the greatest velocity of the particle was 4 and the smallest velocity was -5. So you can imagine 4 at the top of the graph and -5 at the bottom.

A velocity of -5 gives you a speed of +5 (speed is always positive i.e. the magnitude of veliocity) which means that in this case the "greatest speed" of the particle would be 5 but the "greatest velocity" would be 4.

EDIT: "Greatest" velocity could be a confusing term to use depending on the question.
It's a ***** of a question, but I think I've finally got it all down. Thank you!
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