# Is my working correct?

A car is advertised as taking $$11.3\ s$$ to reach a speed of $$100\ kmh^{-1}$$. About how far does it travel before reaching this speed?

Firstly I have tried to work out the acceleration: $$a=\frac{\frac{100000}{3600}}{11.3}\approx 0.2175\ ms^{-2}$$

I've then used this to find the displacement: $$x=0.2175\times 11.3^2\approx 21.7\ m$$

Am I doing something wrong? It seems like too short a displacement...

Check your conversion from $$\tfrac{kilometers}{hour}$$ to $$\tfrac{meters}{second}$$.
Remember that for motion with constant acceleration, starting at rest, $$S=\tfrac{1}{2} a t^2$$
Don't round your results so much. Keep your answer parametric until the very end. This is a skill you'll NEED once you get to the more complicated stuff.
You were off by a factor VERY different from $$10^n$$ in your calculation of the average acceleration.

Another thing I feel I should mention is that you need to state your assumption that the car accelerates with a constant acceleration, as that is not always the case, and your results will vary greatly should you consider such a case.

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I'm now getting the answer as $$\approx 314\ m$$ with
$$a=\frac{2500}{1017}\ ms^{-2}$$
Is that correct?

Last edited:
I'm now getting the answer as $$\approx 314\ m$$ with
$$a=\frac{2500}{1017}\ ms^{-2}$$
Is that correct?

That is very close, you forgot the factor of $$\tfrac{1}{2}$$ in the displacement as a function of time formula.

Oh yeah. Oops.

Thanks again for your help. I'm sure these will become really easy after I do a few more of them.

Oh yeah. Oops.

Thanks again for your help. I'm sure these will become really easy after I do a few more of them.

Oh yeah, completely. By the time you get to the subject of work and energy, you'll be reciting these in your sleep. :)

And no problem, you're very welcome.