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motion ques, help!! (1 Viewer)

jimmik

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a particle with displacement x metres moves in a straight line with velocity v m/s and acceleration a m/s/s. it is given that a = e^(2x - 1) and that at time t = 0 seconds, x = 1/2 and v = -1.

a) show that x = 1/2 - ln(t +1)
b) as t increases indefinitely from 0, what is the range of values for velocity?

any help would be greatly appreciated!
 

CM_Tutor

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You have d<sup>2</sup>x/dt<sup>2</sup> = e<sup>2x-1</sup>
Start by rewriting d<sup>2</sup>x/dt<sup>2</sup> as d/dx(v<sup>2</sup>/2), and integrate with respect to x. Evaluate the constant.
You should now have v<sup>2</sup> = e<sup>2x-1</sup>. Decide whether it follows that v = e<sup>x-1/2</sup> or v = -e<sup>x-1/2</sup>
Rewrite v as dx/dt, manipulate until you have an equation you can integrate, do so and then evaluate the constant. You should have an equation in x and t. Make x the subject, and you should have x = (1 / 2) - ln(t + 1), as required.

As for the last bit, figure out what happens to x as t increases, then use your equation for v in terms of x to figure out what happens to v.
 

jimmik

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i get v = -e^(x-1/2) and integrating this i get t = integ 1/e^(x-1/2) dx. how do u integrate this?
 

CM_Tutor

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It should be t = int -1 / e<sup>x-1/2</sup> dx = - int (e<sup>x-1/2</sup>)<sup>-1</sup> dx = - int e<sup>1/2-x</sup> dx
and you should be able to go on from there.
 

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