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May 16th, 2017, 07:17 AM   #1
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Find acceleration ... velocity is t sin 3t ms^-1

Hi new here. stuck on this one: Find the acceleration of a particle at time "t" seconds given that the velocity is t sin 3t ms^-1.
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May 16th, 2017, 08:14 AM   #2
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Hi new here. stuck on this one: Find the acceleration of a particle at time "t" seconds given that the velocity is t sin 3t ms^-1.
$\dfrac{d}{dt} \bigg[v = t\sin(3t) \bigg]$

product rule ...

$a = \dfrac{dv}{dt} = t \cdot 3\cos(3t) + \sin(3t) \cdot 1$
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May 16th, 2017, 09:45 AM   #3
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calculus acceleration

Thank you for the help; much appreciated.
One more question.....
A particle starting from rest moves with velocity v ms^-1 where v=3t(t-4) find the range of values of t for which acceleration is positive.

3t-4*t is this the start?

Last edited by skipjack; May 17th, 2017 at 06:24 AM.
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May 16th, 2017, 10:29 AM   #4
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Originally Posted by DomB View Post
Thank you for the help; much appreciated.
One more question.....
A particle starting from rest moves with velocity v ms^-1 where v=3t(t-4) find the range of values of t for which acceleration is positive.

3t-4*t is this the start?
No ...

$v = 3t(t-4) = 3t^2 - 12t$

$a = 6t - 12 > 0$

solve the inequality for $t$ ...
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Last edited by skipjack; May 17th, 2017 at 06:24 AM.
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May 16th, 2017, 10:39 AM   #5
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Thank you so much for the help; my calculus is very poor and needs much work. Really appreciate your taking the time to help.

Last edited by skipjack; May 17th, 2017 at 06:18 AM.
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