motion ques, help!! (1 Viewer)

[jimmik]

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]

You have d²x/dt² = e^2x-1
Start by rewriting d²x/dt² as d/dx(v²/2), and integrate with respect to x. Evaluate the constant.
You should now have v² = e^2x-1. Decide whether it follows that v = e^x-1/2 or v = -e^x-1/2
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]

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]

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

 

[jimmik]

ok then thanks