- #1

Yura

- 39

- 0

i got a sheet full of a lot of questions on integrations and differentiations and got them all except for two of them. in both the questions i run into the same type of problem so if i get one then i can probably work out the other using the same method.

here's the question that's troubling me:

(i've worked out part a) but i can't seem to get out part b). )

the part a) answer that i got was 9m/s.

for the part b) its like I am missing some information but seeing as there's two questions like that on the sheet I am guessing there's a method to do this. what i think i need is a function of the Force to x so that i can use

dx/dt = dx/dF * dF/dt.

but unless i have that infomation, right now I am clueless on how to continue.

EDIT: I've done a little more on it but i ended up with the integral of

(-62.5x^2 + 250x)^(-1-2) dx ... but i don't know how to do this integration.

much appreciated if someone can show me how this is done.

thanks.

here's the question that's troubling me:

(i've worked out part a) but i can't seem to get out part b). )

**a particle with mass 80g is acted on by a force which decreases uniformly with respect to displacement from 10N to zero over 2 metres.**

a) calculate the maximum velocity of the particle, given v(0) = 0.

b) find the time for which the force becomes zero.a) calculate the maximum velocity of the particle, given v(0) = 0.

b) find the time for which the force becomes zero.

the part a) answer that i got was 9m/s.

for the part b) its like I am missing some information but seeing as there's two questions like that on the sheet I am guessing there's a method to do this. what i think i need is a function of the Force to x so that i can use

dx/dt = dx/dF * dF/dt.

but unless i have that infomation, right now I am clueless on how to continue.

EDIT: I've done a little more on it but i ended up with the integral of

(-62.5x^2 + 250x)^(-1-2) dx ... but i don't know how to do this integration.

much appreciated if someone can show me how this is done.

thanks.

Last edited: