I cant understand how to prove this question. We learned about intermediate value theorem but this makes no sense because $120$ km isn't in bounds of either upper or lower limit. Here is the question

At $2:00$ PM, a car's speedometer reads $30$km/h. At $2:10$ PM, it reads $50$ km/h. Show that at some time between $2:00$ and $2:10$ the acceleration was exactly $120$ $\text{km}/\text{h}^2$ (YES she wrote the question are $\text{h}^2$. I hope its a typo). Indicate which theorem you must use in your explanation.

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    $\begingroup$ Hint: neither the intermediate value theorem nor the extreme value theorem is the theorem you want. $\endgroup$ Commented Dec 1, 2011 at 13:22
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    $\begingroup$ km/hr^2 is a correct unit for acceleration (velocity per time). $\endgroup$ Commented Dec 1, 2011 at 13:44
  • $\begingroup$ @MikaelÖhman ... per time, i.e., velocity per time per time. $\endgroup$
    – GeoffDS
    Commented Dec 1, 2011 at 14:13
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    $\begingroup$ Let's say that you've got 2/3rds of the name of the theorem you need correct. $\endgroup$ Commented Dec 1, 2011 at 14:18
  • $\begingroup$ @Grapth: Mikael is correct: $$\text{velocity} = \frac{\text{displacement}}{\text{time}},$$ while $$\text{acceleration} = \frac{\text{velocity}}{\text{time}} = \frac{\text{displacement}}{(\text{time})^2}$$ $\endgroup$
    – JavaMan
    Commented Dec 1, 2011 at 14:52

2 Answers 2


First, convert the units of your time axis to hours: say 2:00 $= 0$, and 2:10 $ = 1/6$. What does the mean value say about the acceleration (= derivative of velocity) for the endpoint values 30 and 50 respectively? I don't want to give too much away.


$$ \frac{50\text{ km}/\text{hr} - 30\text{ km}/\text{hr}}{1/6\text{ hr}} = \frac{20}{1/6} \ \frac{\text{km}}{\text{hr}^2}. $$ There's no typo; acceleration can be measured in kilometers per hour per hour, written as $\text{km}/\text{hr}^2$, or in meters per second per second, written as $\text{m}/\text{sec}^2$, etc.

Acceleration is the derivative of velocity with respect to time. So you have velocity at one point in time, minus velocity at another point in time, divided by elapsed time, equals the derivative of velocity with respect to time at some time between those two points in time. A standard theorem deals with that.

  • $\begingroup$ what theorem is it, mean value theorem :S? $\endgroup$
    – Raynos
    Commented Dec 1, 2011 at 15:56
  • $\begingroup$ @Nadal : It's the mean value theorem. The number $20/(1/6)=120$ is the average (or mean) acceleration during those 10 minutes. The mean value theorem says that there is some point in time between the two times the instantaneous acceleration is equal to the average acceleration over the whole time interval. $\endgroup$ Commented Dec 2, 2011 at 16:21

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