Question

Acceleration of a particle is given by $a=(2+\frac{5t}{2})m/s^2$. At $t=0$, velocity of the particle is $3m/s$ and its position is $x=\frac{5}{3}m$. Find the value of $\frac{18}{5}\left(\frac{V_{\text{avg}}}{V_{\text{inst}}}\right)$, where $V_{\text{avg}}$ is average velocity for $t=0$ to $2s$ and $V_{\text{inst}}$ is the instantaneous velocity at $t=2s$.

Answer 1

Step 1: Find the velocity of the particle.
Formula used: $a=\frac{dv}{dt}$
Given:
$a=(2+\frac{5t}{2})m/s^2$
At $t=0,u=3m/s,x=\frac{5}{3}m$
As we know,
$a=\frac{dv}{dt}adt=dv$

Integrating both sides,
${\int }_0^{t}adt={\int }_0^{v}dv{\int }_0^t(2+\frac{5t}{2})dt={\int }_0^{v}dv$
Therefore, velocity of particle is given by $v=2t+\frac{5t^2}{4}+3$
At $t=2s,v=12m/s$.

Step 2: Find the displacement of the particle.
Formula used: $v=\frac{dx}{dt}$
As we know, $v=\frac{dx}{dt}vdt=dx$
Integrating both sides, ${\int }_0^{t}vdt={\int }_0^{x}dx{\int }_0^t(2t+\frac{5t^2}{4}+3)dt=x$
Therefore,
$x=t^2+\frac{5t^3}{12}+3t$
At $t=0,x=0$
At $t=2,x=\frac{40}{3}m/s$.

Step 3: Find the average velocity for $t=0$ to $2s$ of the particle.
$V_{\text{avg}}=\frac{x\left(2\right)-x\left(0\right)}{2}=\frac{20}{3}m/s$

Step 4: Find the instantaneous velocity ar $t=2s$ of the particle.
$V_{\text{inst}}=V\left(2\right)=12m/s$

Final answer: $V_{\text{avg}}=\frac{20}{3}m/s$, $V_{\text{inst}}=12m/s$.