Question

The motion of a particle is defined by the equations $x=4t^2$ and $y=2t^3$ in $meters$, where $t$ is in $seconds$. Determine the normal and tangential components of the particle’s velocity and acceleration when $t=2s$

Answer 1

Step 1: Given Data:

Here, $r=\left(4t^2\right)\hat{i}+\left(2t^3\right)\hat{j}$

Now we need to differentiate $r$ with respect to $t$,

$v=dr/dt=\left(8t\right)\hat{i}+\left(6t^2\right)\hat{j}$

When $t=2s,v=(8\times 2)\hat{i}+(6\times 2^2)\hat{j}=16\hat{i}+24\hat{j}$

Then, $$\begin{gathered} v=\sqrt{{16}^2+{24}^2} \\ v=\sqrt{256+576}=\sqrt{832} \\ v=28.8m/s \end{gathered}$$

From $v$ value we can say that velocity is directed tangent to the given path.

Now $v_n=0$ and $v_t=28.8m/s$

Then the velocity $v$ makes an angle $\theta ={\tan }^{-1}(24/16)=56.{31}^{°}$

Step 2 : Calculate acceleration

Accereration, $a=dv/dt=8\hat{i}+12\hat{j}m/s^2$

at $$\begin{gathered} t=2s,a=8\hat{i}+12\left(2\right)\hat{j}=8\hat{i}+24\hat{j} \\ a=\sqrt{8^2+{24}^2}=25.298 \end{gathered}$$

Now here acceleration a makes an angle $\varphi ={\tan }^{-1}(24/8)={\tan }^{-1}\left(3\right)=71.56°$

$\alpha =71.560-56.310=15.25°$

$$\begin{gathered} a_n=a\sin \alpha =25.298\sin 15.25=6.585m/s^2 \\ {a}_t=a\cos \alpha =25.298\cos 15.25=24.404m/s^2 \end{gathered}$$

Hence the normal and tangential components of the particle’s velocity and acceleration when $t=2s$ are $v_n=0m/s$, $v_t=28.8m/s$ and $a_n=6.585$, $a_t=24.404$.