Calculus made easy exercise question Exercise 5 Question 10.

Question

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.

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}

Then,

\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.