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mechanics help needed pls

Not sure how to find the greatest acceleration for model Q (part v)
m1.PNG
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m2.PNG50.7KB
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h26
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for model p work out gradient of steepest line on graph, for model q differentiate the equation
for Q you need to find dv/dt and equate to zero.

since the derivative is linear there is no ambiguity about whether it is a max or a min.
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Original post by ben288888
for model p work out gradient of steepest line on graph, for model q differentiate the equation

ok so Ive already done both. But what do you do after differentiating model q equation?
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Original post by the bear
for Q you need to find dv/dt and equate to zero.

since the derivative is linear there is no ambiguity about whether it is a max or a min.

iam confused so why do you equal it to zero?
Original post by h26
iam confused so why do you equal it to zero?


sorry my bad. you should differentiate the acceleration and equate it to zero.
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Original post by the bear
sorry my bad. you should differentiate the acceleration and equate it to zero.

that doesnt make sense either so differentiate a to get -1/4 and then make -1/4 equal to 0 :s-smilie:

ms
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(edited 5 years ago)
Original post by h26
that doesnt make sense either so differentiate a to get -1/4 and then make -1/4 equal to 0 :s-smilie:

ms
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so a = 2.5 - 0.25t

you are right there is no local max or local min. just put in a suitable value of t from the range 0 to 12 seconds to find the largest acceleration.
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Original post by h26
that doesnt make sense either so differentiate a to get -1/4 and then make -1/4 equal to 0 :s-smilie:

ms
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Differentiation gives 5/2 - (1/4)t
Original post by the bear
so a = 2.5 - 0.25t

you are right there is no local max or local min. just put in a suitable value of t from the range 0 to 12 seconds to find the largest acceleration.


Put in t = 0 as anything above this will subtract from the initial 2.5, so you cannot get higher than 2.5
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Original post by the bear
so a = 2.5 - 0.25t

you are right there is no local max or local min. just put in a suitable value of t from the range 0 to 12 seconds to find the largest acceleration.

Thanks!
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Original post by Joshey
Differentiation gives 5/2 - (1/4)t


Put in t = 0 as anything above this will subtract from the initial 2.5, so you cannot get higher than 2.5

Thanks!!

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