This an M2 practice question that I can't wrap my head around. Any help or advice would be greatly appreciated.


A car is travelling along a straight horizontal road against resistances to motion which are constant and total $200N$. When the engine of the car is working at a rate of $H$ kilowatts, the maximum speed of the car is $30\;ms^ {-1}$.

a) Find the value of H.

I could do this bit and got the correct value of 72. No problems there.

The driver wishes to overtake another vehicle so she increases the rate of working of the engine by $20$% and this results in an initial acceleration of $0.32\;ms^{-2} $. Assuming that the resistances to motion remain constant,

b) Find the mass of the car.

For this section I understand to do $\frac{power}{speed} -2000 = m\times 0.32 $, and I can work out the power as $72000$. However, I don't understand how I can assume that the speed is still $30 ms^{-1}$.

  • $\begingroup$ You are only looking at the initial moment. When the driver starts the acceleration. $\endgroup$ – Steeven Jun 12 '18 at 21:03

I don't understand how I can assume that the speed is still $30\ ms^{-1}$

The magic word here is the word "initial":

The power of the engine is calculated as:

$P = (F_{resistance} + ma)v$

Or (assuming the power and the resistance are constant):

$a(t) = (Pv^{-1}(t) - F_{resistance})m^{-1}$

This means that the acceleration will not be constant but it is depending on the speed of the car.

Now you have to keep in mind what differential calculus actually is:

We are looking at the time span $t=0...1s$ to get a rough result, we get a better result when looking at the time span $t=0...0.1s$ and the result will even be better when looking at the time span $t=0...0.01s$ and so on...

The shorter the time span is the better is the result.

When looking at the time span ${t=0...t_n}$ we know that the speed of the car is somewhere in the range $30ms^{-1} \leq v < 30ms^{-1}+0.32ms^{-2}*t_x$.

The word "initial" means that $0.32ms^{-2}$ is the acceleration immediately after increasing the power of the engine.

And the shorter the time span we are looking at is the better is our result.

So when we choose a $t_x$ very close to 0 the result should be the best. And you can see that in this case the inequation above will look like this:

$30ms^{-1} \leq v < 30ms^{-1}+0.32ms^{-2}*(nearly\ 0)$

Or (because 0*x = 0):

$30ms^{-1} \leq v < 30ms^{-1}+(nearly\ 0)$

Or (because 0+x = x):

$30ms^{-1} \leq v < (nearly\ 30ms^{-1}$)


Calculate the rate of change of kinetic energy at the first moment of acceleration from the fact that $KE = (m v^2)/2 => d/dt (KE) = m v a$, which is equal to the increased power of the engine. You can relate that to the rate of change of momentum, which equals mass times acceleration, then solve for mass.


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