Motion and differentiation question (1 Viewer)

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Hey guys, this probably isn't very hard but I'm very confused! Any help would be much appreciated
A particle is moving in a straight line so that its displacement xcm over time t seconds is given by x=t(square root: 49-t²)
How far does the particle move altogether?
Sorry for the confusion about the square root sign, basically just the 49 - t² is all under a square root multiplied by t on the outside!
Thankya

 

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somebody help please!!

 

[braintic]

When you differentiate and factorise, you should get v = (49-2t^2) / sqrt(49-t^2)

This is equal to zero at t = 7/sqrt(2)

The particle starts at x=0, and ends its motion 7 seconds later also at x=0 (due to the domain restriction).

So you need to substitute t=7/sqrt(2) into the function and double.

But ..... where did you get this question from? It is absolutely ridiculous.
Apparently the particle goes from an extremely high negative velocity just before t=7, to just stopping dead.
I'm guessing it must have been annihilated at t=7 by a beam of anti-matter.

(Not to mention the fact that the particle exceeds the speed of light)

 

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Thankyou but I still don't quite understand! Why wouldn't you sub in t=7 because its asking how far the particle has travelled and altogether it travels for 7 seconds, not 7/sqrt(2) ? Its just from some random sheet my maths teacher gave me ahah

 

[braintic]

Thankyou but I still don't quite understand! Why wouldn't you sub in t=7 because its asking how far the particle has travelled and altogether it travels for 7 seconds, not 7/sqrt(2) ? Its just from some random sheet my maths teacher gave me ahah

I did sub in t=7 ... as I said in the 3rd line, I got x=0.

So it starts at x=0 and ends at x=0. But that doesn't tell you how far its travelled .... you need to know where its been during that 7 seconds.
It turned around at t=7/sqrt(2) because that it where it was stationary, so it started at x=0, headed to whatever x value at t=7/sqrt(2), then returned to x=0 at t=7.

 

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Ok I understand now, thanks so much