1. distance travelled

a car travelling in a straight line experiences a retardation which is proportional to its speed cubed. the speed of the car reduces from 20 to 10 m/s^-1 in a time of t seconds
show that the distance travelled in this time is 40t/3 m

so what I tried to do was let dv/dt =-kv^3 I put in the limits and I got k=1/40t

I then let vdv/ds =-kv^3 and I put in the limits and I got 1/20 =ks and I subbed in k and i got the wrong answer.

2. Re: distance travelled

This looks pretty straightforward

$\dot{v} = k v^3$

$\dfrac{dv }{v^3} = k~dt$

$-\dfrac{1}{2v^2} = k t + v_0$

$v(t) = \dfrac{1}{\sqrt{2(k t+v_0)}}$

$v(0) = 20,~v(t)=10$

$k = \dfrac{3}{800t},~v_0 = \dfrac{1}{800}$

$\displaystyle \int_0^t~v(\tau)~d\tau = \dfrac{40t}{3}$

I've given you just the skeleton obviously, you have to fill in the meat, but the problem is legit.

3. Re: distance travelled

Originally Posted by edwardkiely
a car travelling in a straight line experiences a retardation which is proportional to its speed cubed. the speed of the car reduces from 20 to 10 m/s^-1 in a time of t seconds
show that the distance travelled in this time is 40t/3 m

so what I tried to do was let dv/dt =-kv^3 I put in the limits and I got k=1/40t
This makes no sense the "constant of proportion" is a constant, not a function of t.

I then let vdv/ds =-kv^3 and I put in the limits and I got 1/20 =ks and I subbed in k and i got the wrong answer.

4. Re: distance travelled

Originally Posted by HallsofIvy
This makes no sense the "constant of proportion" is a constant, not a function of t.
terrible use of notation but in this case $t$ is a constant.

it's the length of time it takes for the car to decelerate from 20 m/s to 10 m/s

5. Re: distance travelled

Thanks, I would have used a capital "T".