# How to find the tangential acceleration when a function is given with a constant as unknown?

#### Chemist116

The problem is as follows:

A crystal is moving on an horizontal plane $x-y$ by the given law: $r(t)=\left(12t\hat{i}+ct^2\hat{j}\right)\,m$ with $t$ being the time on seconds and $c$ a positive constant with given acceleration units. If for t=0 the radius of curvature is $4\,m$. Find the tangential acceleration for $t=2\,s$.

The given alternatives are:

$\begin{array}{ll} 1.&\frac{216}{37}\sqrt{37}\,\frac{m}{s^2}\\ 2.&216\sqrt{37}\,\frac{m}{s^2}\\ 3.&\frac{\sqrt{37}}{37}\,\frac{m}{s^2}\\ 4.&\frac{72}{37}\sqrt{37}\,\frac{m}{s^2}\\ 5.&\frac{144}{37}\sqrt{37}\,\frac{m}{s^2}\\ \end{array}$

I'm confused exactly how to tackle this problem:

It seems obvious that it is needed the value of $c$ because with that, then I could obtain an expression from where it can be taken its derivative consecutively and with that the acceleration.

But the thing is if I do plug in the initial condition from t=0

The whole equation becomes zero.

$r(t)=(12t\hat{i}+ct^2\hat{j})$

$r(0)=(12t\hat{i}+ct^2\hat{j}) = 0$

So what can be done here?.

I cannot assume that the radius of curvature $4$ will be the same for $t=2$.

Can somebody help me here?.

Last edited:

#### skeeter

Math Team
$r(t) = x(t) \hat{i} + y(t) \hat{j}$ is the position of the object in the x-y plane as a function of time.

radius of curvature as a function of time is $R(t) = \dfrac{\left([x'(t)]^2 + [y'(t)]^2\right)^{3/2}}{|x'(t)y''(t) - y'(t)x''(t)|}$

you are given $x(t) = 12t$ and $y(t)=ct^2$ ... from these, the necessary derivatives can be found in order to determine the constant $c$ from the given $R(0) = 4$

Once you find $c$, you can further determine $R(2)$

Finally ...

$a = \sqrt{a_T^2 + a_c^2} \implies a_T = \sqrt{a^2 - a_c^2}$, where $a = \sqrt{[x''(t)]^2+[y''(t)]^2}$ and $a_c = \dfrac{v^2}{R} = \dfrac{[x'(t)]^2+[y'(t)]^2}{R}$

topsquark