Math Help - 1st & 2nd Derivative Physics word problem

1. 1st & 2nd Derivative Physics word problem

I'm stuck on this problem. Can someone help me with where to start?

A body moves in such a way that the space described in the time t is given by $s=t^n$, where n is a constant. Find the value of n when the velocity is doubled from the 5th to the 10th second; find it also when the velocity is numerically equal to the acceleration at the end of the 10th second.

2. Hello, dbakeg00!

You're expected to know that: . $\begin{array}{ccc}v &=& \dfrac{ds}{dt} \\ \\[-3mm] a &=& \dfrac{dv}{dt} \end{array}$

A body moves in such a way that its position at time $t$ is given by: . $s\:=\:t^n$, where n is a constant.

(a) Find $n$ when the velocity is doubled from the 5th to the 10th second.
We have: . $v(t) \:=\:n\,t^{n-1}$

Then: . $\begin{Bmatrix}v(5) &=& n\,5^{n-1} \\ v(10) &=& n\,10^{n-1} \end{Bmatrix}$

$v(10)$ is twice $v(5)\!:\;\;n\,10^{n-1} \:=\:2\cdot n\,5^{n-1} \quad\Rightarrow\quad \frac{10^{n-1}}{5^{n-1}} \:=\:2 \quad\Rightarrow\quad 2^{n-1} \:=\:2^1$

Therefore: . $n-1 \:=\:1 \quad\Rightarrow\quad n \:=\:2$

(b) Find $n$ when the velocity is numerically equal to the acceleration at the end of the 10th second.
We have: . $\begin{Bmatrix}v(t) &=& n\,t^{n-1} \\ a(t) &=& n(n-1)t^{n-2} \end{Bmatrix}$

Then: . $\begin{Bmatrix} v(10 &=& n\,10^{n-1} \\ a(10) &=& n(n-1)\,10^{n-2} \end{Bmatrix}$

These two quantities are equal: . $n(n-1)\,10^{n-2}\;=\;n\,10^{n-1}$

Divide by $n\,10^{n-2}\!:\;\;n-1 \:=\:10$

. . . . . Therefore: . $n \:=\: 11$

3. Thanks for the quick reply and for your answer, it helped a lot. One more thing...

These two quantities are equal: . $n(n-1)\,10^{n-2}\;=\;n\,10^{n-1}$

Divide by $n\,10^{n-2}\!:\;\;n-1 \:=\:10$

. . . . . Therefore: . $n \:=\: 11$
So when you divide $n\,10^{n-1}$ by $n\,10^{n-2}$, how do you come up with 10 ?(sorry, I'm a little rusty on division by exponents..I know you should subtract them, but I am having a difficult time seeing this one)

4. Nevermind...I'm an idiot! I see it now. Thanks again for the help.