Answer to Question #155069 in Classical Mechanics for Expo

Question #155069
  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.
1
Expert's answer
2021-01-12T12:10:33-0500

Given,

Mass of the particle (m) = 2kg

Particle starts from t = 0

Force F=2i^+4tj^+6t2k^F= 2 \hat{i} + 4t\hat{ j} + 6t^2\hat{ k}

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

Hence, acceleration of the particle (a)=2i^+4tj^+6t2k^2(a)=\frac{2 \hat{i} + 4t\hat{ j} + 6t^2\hat{ k}}{2}

a=i^+2tj^+3t2k^\Rightarrow \overrightarrow{a} = \hat{i} + 2t\hat{ j} + 3t^2\hat{ k}

a=1+4t2+9t4|a|=\sqrt{1+4t^2+9t^4}

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

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

Now, substituting the values, dv=t=0t(i^+2tj^+3t2k^)dt\int dv= \int_{t=0}^{t}(\hat{i} + 2t\hat{ j} + 3t^2\hat{ k})dt

dv=(ti^+122t2j^+133t3k^)\Rightarrow \int dv= (t\hat{i} + \frac{1}{2}2t^2\hat{ j} +\frac{1}{3}3t^3\hat{ k})

v=(ti^+t2j^+t3k^)v= (t\hat{i} + t^2\hat{ j} +t^3\hat{ k})

v=t2+t4+t6|v|=\sqrt{t^2+t^4+t^6}

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

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

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

Now, substituting the values,

dx=t=0t(ti^+t2j^+t3k^)dt\Rightarrow \int \overrightarrow{dx}=\int_{t=0}^{t}(t\hat{i} + t^2\hat{ j} +t^3\hat{ k})dt

x=(t22i^+t33j^+t44k^)\Rightarrow \overrightarrow{x}=(\frac{t^2}{2}\hat{i} + \frac{t^3}{3}\hat{ j} +\frac{t^4}{4}\hat{ k})

x=t44+t69+t816|x|=\sqrt{\frac{t^4}{4}+\frac{t^6}{9}+\frac{t^8}{16}}


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