Question

1. A particle of mass m =2kg starts from rest at the origin of an integral coordinate system at time t=0. A force F = 2 i + 4t j + 6t2 k is applied to it. Find the acceleration, velocity, and position of the particle for any latter time.

Answer 1

Given,

Mass of the particle (m) = 2kg

Particle starts from t = 0

Force $F= 2 \hat{i} + 4t\hat{ j} + 6t^2\hat{ k}$

Acceleration of the particle $a=\frac{F}{m}$

Hence, acceleration of the particle $(a)=\frac{2 \hat{i} + 4t\hat{ j} + 6t^2\hat{ k}}{2}$

$\Rightarrow \overrightarrow{a} = \hat{i} + 2t\hat{ j} + 3t^2\hat{ k}$

$|a|=\sqrt{1+4t^2+9t^4}$

we know $\overrightarrow{a}=\frac{dv}{dt}$

Hence, $\overrightarrow{dv}=\overrightarrow{a}dt$

Now, substituting the values, $\int dv= \int_{t=0}^{t}(\hat{i} + 2t\hat{ j} + 3t^2\hat{ k})dt$

$\Rightarrow \int dv= (t\hat{i} + \frac{1}{2}2t^2\hat{ j} +\frac{1}{3}3t^3\hat{ k})$

$v= (t\hat{i} + t^2\hat{ j} +t^3\hat{ k})$

$|v|=\sqrt{t^2+t^4+t^6}$

We know that the relation between the position and the velocity is

$\overrightarrow{v}=\frac{\overrightarrow{dx}}{dt}$

Now, $\overrightarrow{dx}=\overrightarrow{v}dt$

Now, substituting the values,

$\Rightarrow \int \overrightarrow{dx}=\int_{t=0}^{t}(t\hat{i} + t^2\hat{ j} +t^3\hat{ k})dt$

$\Rightarrow \overrightarrow{x}=(\frac{t^2}{2}\hat{i} + \frac{t^3}{3}\hat{ j} +\frac{t^4}{4}\hat{ k})$

$|x|=\sqrt{\frac{t^4}{4}+\frac{t^6}{9}+\frac{t^8}{16}}$