A body moves in such a way that the spaces described in the time $t$ from starting is given by $s = t^n$ ,where $n$ is a constant.

  1. Find the value of $n$ when the velocity is doubled from the $5^{th}$ to the $10^{th}$ second

  2. find it also when the velocity is numerically equal to the acceleration at the end of the 10th second.

How do I find answer to this question?
This is an exercise question in Calculus made easy. Do I need to know logarithms to find the answer? v=dS/dt = nt^(n-1) a = dV/dt = n(n-1)*t^(n-2). I wasn't able to understand what to do with these though.The answers are n=2 and n =11 respectively.When I tried to approach the question I was stuck on solving for X in exponent.

  • $\begingroup$ Spaces described? What does that mean? $\endgroup$ Commented May 21, 2015 at 16:07
  • $\begingroup$ What is it that you doubt? $\endgroup$
    – WillO
    Commented May 21, 2015 at 16:17
  • $\begingroup$ @ThomasAndrews I am not sure, most probably the space covered by the body. $\endgroup$
    – Prnv
    Commented May 21, 2015 at 16:21
  • $\begingroup$ @WillO I want to know how to approach this question. $\endgroup$
    – Prnv
    Commented May 21, 2015 at 16:21
  • $\begingroup$ @Math: A good first step might be to find out what the question means. $\endgroup$
    – WillO
    Commented May 21, 2015 at 16:24

1 Answer 1


Interpreting "spaces" to mean distance, we have that velocity is how much the distance changes in a given time. Therefore, we need to differentiate $s$ with respect to $t$. Therefore,

\begin{align} \frac{ds}{dt}=nt^{n-1}\,. \end{align}

Calling this result, $v(t)$, we want to find the value of $n$ if we know

\begin{align} \frac{v(10)}{v(5)}=2\,. \end{align}


\begin{align} 2&=\frac{v(10)}{v(5)}=\frac{n10^{n-1}}{n5^{n-1}}=\frac{10^{n-1}}{5^{n-1}}\\&=\frac{10^n10^{-1}}{5^n5^{-1}}=\frac{10^{-1}}{5^{-1}}\times\frac{10^n}{5^n}=\frac{\frac{1}{10}}{\frac{1}{5}}\times\left(\frac{10}{5}\right)^n=\frac{5}{10}\left(\frac{10}{5}\right)^n\\&=\frac{1}{2}\left(\frac{10}{5}\right)^n\,. \end{align}

As a result, we have

\begin{align} 2=\frac{1}{2}2^n\,. \end{align}

This gives us

\begin{align} 4=2^n. \end{align}

The question now becomes: "What power of $2$ gives us $4$?" You can use logarithms if you want, or you might remember that the answer is $n=2$.

To find the acceleration, we differentiate the velocity, \begin{align} \frac{ds}{dt}=nt^{n-1}\,, \end{align} with respect to $t$. The process is similar, but now you set the first and second derivatives equal to each other (with $t=10$) and solve for $n$. What I believe you'll find is that $10^n$ will cancel, so you shouldn't need any logarithms to solve.

  • $\begingroup$ Thanks, How did 10^(n-1)/5 ^(n-1) turn to 1/2 * (10/5)^n though? $\endgroup$
    – Prnv
    Commented May 21, 2015 at 16:37
  • $\begingroup$ @Math I added an extra line in the third set of equations. I hope it clarifies things. I essentially broke up the $\frac{10^{n-1}}{5^{n-1}}$ and simplified. $\endgroup$
    – user85503
    Commented May 21, 2015 at 16:43
  • $\begingroup$ Thanks for the extra steps,I tried this with the second sub question as well and it worked :) $\endgroup$
    – Prnv
    Commented May 21, 2015 at 17:03
  • $\begingroup$ Great! Glad I could help. $\endgroup$
    – user85503
    Commented May 21, 2015 at 17:08

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