diffentiating- for acceleration Watch

WWEKANE
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Hi. Is it ok if you can help me with this question I have


I calculated gradient in 2 ways . One by diffentiating and then using the second derivative to calculate the acceleration at that time.


Another by using the calculating change in velocity over time


I get 2 different answers


Surley acceleration between 1 and 3 seconds must be constant (I think 12 ms^-2)


As this means that at 2 seconds it will reach 15 ms-1 and at 3 seconds it would then reach 27ms-1.* So acceleration must be constant to reach that speed(27) after 3 seconds*


Which answer is correct and can you only use the second method if the range was 1.5 seconds to 3.5 Seconds.*


Thank you for your time*


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old_engineer
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There's not enough information given in the question (as quoted) to arrive at a meaningful answer. the car could be travelling at 3m/s at t = 1, then stop completely, then accelerate rapidly to 27m/s just before t = 3. Have you posted the complete question?
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WWEKANE
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it states the function of s=t^3 isn't this enough to show it doesn't stop rapidly?
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RogerOxon
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(Original post by WWEKANE)
Hi. Is it ok if you can help me with this question I have


I calculated gradient in 2 ways . One by diffentiating and then using the second derivative to calculate the acceleration at that time.


Another by using the calculating change in velocity over time


I get 2 different answers


Surley acceleration between 1 and 3 seconds must be constant (I think 12 ms^-2)


As this means that at 2 seconds it will reach 15 ms-1 and at 3 seconds it would then reach 27ms-1.* So acceleration must be constant to reach that speed(27) after 3 seconds*


Which answer is correct and can you only use the second method if the range was 1.5 seconds to 3.5 Seconds.*


Thank you for your time*


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Where did you get your equation for s from?

If your equations are correct, this is not constant acceleration, so you cannot calculate acceleration from the change in speed over finite time - you must use calculus.

Can you post the full question?
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old_engineer
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Ah, OK
(Original post by WWEKANE)
it states the function of s=t^3 isn't this enough to show it doesn't stop rapidly?
Ah, OK, if s = t^3 is part of the question then there is enough information given. It looked like part of your working. Your second method, based on deltaV / t, would work if the acceleration was constant over time, but it breaks down when acceleration varies with time.
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WWEKANE
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I made this question up to try understand why my answers were different. I am mainly confused as to why acceleration isn't constant as surley it needs to travel at 12 ms-1 per second to reach that speed at 3 seconds when using s=t^3.
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WWEKANE
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oh ok. so would the change in velocity compensate each other so at the end of 3 seconds the velocity reached is still 27
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sotor
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the acceleration is not constant, if s=t^3 then acceleration=6t which means it is constantly changing as time changes! therefore you have to do calculus. dont even think about another method in the exam, if its variable acceleration you have to automatically think calculus.

but also, 12m/s^2 is the same as 6(2)m/s^2. you got a=6t from differentiation

you answers are different because your "change over time" answer is the acceleration at t=2, and your calculus answer is the acceleration specifically at the point t=2.5s
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WWEKANE
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ok thank you. so the acceleration is only the same for time = 2? for both methods why is this the case. thanks for you detailed response
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sotor
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(Original post by WWEKANE)
ok thank you. so the acceleration is only the same for time = 2? for both methods why is this the case. thanks for you detailed response
i think it's just chance to be honest.

by doing change over time i think you worked out the average acceleration, not the acceleration at a point. its possible that this just happened to also be the acceleration at 2.
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