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Old Oct 22nd 2017, 06:53 AM   #1
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Motion in 1d

A particle moves in a straight line and itís position x is at time t is given by x^2 = 2 + t . Find acceleration
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Old Oct 22nd 2017, 07:31 AM   #2
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Differentiate both sides of that equation with respect to time to find the velocity:

The derivative of x^2 with respect to t is 2x dx/dt and the derivative of 2+ t is 1. 2x dx/dt= 2xv= 1.

Differentiate again to find the acceleration.

Using the product rule, the derivative of 2x v is 2 v^2+ 2x dv/dt= 2v^2+ 2xa. Of course the derivative of 1 is 0.

So we have 2v^2+ 2xa= 0. From 2xv= 1 , v= 1/(2x) so we can write 2(1/(4x^2))+ 2x a= 1/(2x^2)+ 2x a= 0. The acceleration is -1/(4x^3).
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Old Oct 22nd 2017, 08:17 PM   #3
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Originally Posted by HallsofIvy View Post
Differentiate both sides of that equation with ...
Apparently HallsofIvy isn't aware of the rules of this forum. We do not give a full answer to a problem posted with no attempt. Otherwise we'd risk doing a students homework forum them.

HoI - Please read the forum rules: Must read before posting a problem : please show your attempt!
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Old Oct 23rd 2017, 06:42 AM   #4
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Sorry I'll show my attempt now onwards
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