M3 - differential equations

Hi,

I was hoping someone could help me with this question:

If a car of mass 1250kg is travelling at v m/s, the resistance is 5v + v^2 newtons (no forwards force as engine is disengaged). If it is initially moving at 20m/s,
a) how long does it take for speed to drop to 5 m/s?
b) how far does it travel while slowing down to 5 m/s?

My working so far:

a) from F=ma,

-5v - v^2 = 1250 \frac{dv}{dx}

Then I guessed that we were separating the variables, so we have:

-\frac{1}{1250}dt = \frac{1}{5v+v^2}dv

I would then like to integrate both sides, but I can't integrate the RHS. Any ideas?

Thanks

Reply 1

DFranklin

18

Do you mean

\frac{dv}{dx}

or latex]\frac{dv}{dt}

For the integral, use partial fractions.

Reply 2

Lifesharker

jess121

Hi,

I was hoping someone could help me with this question:

If a car of mass 1250kg is travelling at v m/s, the resistance is 5v + v^2 newtons (no forwards force as engine is disengaged). If it is initially moving at 20m/s,
a) how long does it take for speed to drop to 5 m/s?
b) how far does it travel while slowing down to 5 m/s?

My working so far:

a) from F=ma,

-5v - v^2 = 1250 \frac{dv}{dx}

Then I guessed that we were separating the variables, so we have:

-\frac{1}{1250}dt = \frac{1}{5v+v^2}dv

I would then like to integrate both sides, but I can't integrate the RHS. Any ideas?

Thanks

You can seperate the RHS by partial fractions;

1/(5v + v^2) = A/(5+v) + B/v

Which you can solve for A and B by multiplying both sides by (5v + v^2) and subbing in values for v.