# Thread: particle moving

1. ## particle moving

the deceleration of a particle moving in a straight line with speed v has magnitude 2e^v .It's initial speed is 10 m/s

(i)find the time t1 for the speed to half in magnitude.
(ii)find the time t2 for the particle to come to rest.
(iii) deduce (t2-t1)/t1 =e^5

I can not do any of this question.

I am letting dv/dt = -2e^v I am having trouble seperating this and I do not know how to integrate it.

2. ## Re: particle moving

$\dfrac{dv}{dt} = 2 e^v$

$e^{-v} dv = 2 dt$

$-e^{-v} = 2t + C$

$e^{-v} = -2t + C$ (constant absorbs the negative sign)

$-v = \ln(-2t+C)$

$v = -\ln(-2t+C)$

$v(0) = -\ln(C) = 10$

$C = e^{-10}$

$v(t) = -ln(-2t + e^{-10})$

$v(t_1) = \dfrac{10}{2} = 5$

$-\ln(-2t_1 + e^{-10}) = 5$

$-2t_1 + e^{-10} = e^{-5}$

$-2t_1 = e^{-5}-e^{-10}$

$t_1 = \dfrac{e^{-10}-e^{-5}}{2} = \dfrac{1-e^5}{2e^{10}}$

see if you can figure out the rest

3. ## Re: particle moving

That's pretty straight forward: dv/dt= -2e^v separates to (e^{-v})dv= -2dt

Since the derivative of e^{ax} is ae^{ax}, the integral of e^{ax} is e^{ax}/a. I presume you can integrate -2dt.

Since the initial speed was 10, v(t1)= 5. Solve that for t1.

v(t2)= 0. Solve that for t2.

4. ## Re: particle moving

thanks.

why does the c absorb the minus and why does it not become -c

5. ## Re: particle moving Originally Posted by edwardkiely thanks.

why does the c absorb the minus and why does it not become -c
Because it's an arbitrary constant. It can be anything at all so using a negative sign is superfluous.