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?.
Your velocity is $v=(12,2ct)$ and your acceleration $a=(0,2c)$. Since $v(0)=(12,0)$ at $t=0$, at that moment all the acceleration is normal (no tangential). Therefore, $$ a(0)=a_N(0)=\frac{v(0)^2}{R(0)}= \frac{144}{4}=36. $$ This implies $2c=36$ or $c=18$. Next, you find $v(2)=(12,72)$. We know $a(2)=(0,36)$. To find the tangential part, just project in the direction of the velocity, $$ a_T(2)=(0,36)\cdot (12,72)/\|v(2)\|=36\cdot 72/\sqrt{5328}=36\cdot 72/12\sqrt{37}=216/\sqrt{37}=216\sqrt{37}/37, $$ which is your option (a).