My Math Forum Find acceleration ... velocity is t sin 3t ms^-1

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 May 16th, 2017, 07:17 AM #1 Newbie   Joined: May 2017 From: uk Posts: 9 Thanks: 0 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.
May 16th, 2017, 08:14 AM   #2
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Quote:
 Originally Posted by DomB 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 ...

$a = \dfrac{dv}{dt} = t \cdot 3\cos(3t) + \sin(3t) \cdot 1$

 May 16th, 2017, 09:45 AM #3 Newbie   Joined: May 2017 From: uk Posts: 9 Thanks: 0 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.
May 16th, 2017, 10:29 AM   #4
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,751
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Quote:
 Originally Posted by DomB 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 $t$ ...

Last edited by skipjack; May 17th, 2017 at 06:24 AM.

 May 16th, 2017, 10:39 AM #5 Newbie   Joined: May 2017 From: uk Posts: 9 Thanks: 0 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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