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Question

Physics Question on projectile motion

The coordinates of a moving particle at any time tt are given by x=αt3x=\alpha t^{3} and y=βt3y=\beta t^{3}. The speed of the particle at time tt is given by

A

3tα2+β23 t \sqrt{\alpha^{2}+\beta^{2}}

B

3t2α2+β23 t^2 \sqrt{\alpha^{2}+\beta^{2}}

C

t2α2+β2t^{2} \sqrt{\alpha^{2}+\beta^{2}}

D

α2+β2\sqrt{\alpha^{2}+\beta^{2}}

Answer

3t2α2+β23 t^2 \sqrt{\alpha^{2}+\beta^{2}}

Explanation

Solution

x=αt3,y=βt3x=\alpha t^{3}, y=\beta t^{3}
vx=dxdt=3αt2v_{x}=\frac{d x}{d t}=3 \alpha t^{2}
vy=dydt=3βt2v_{y}=\frac{d y}{d t}=3 \beta t^{2}
Resultant velocity v=vx2+vy2v= \sqrt{v_{x}^{2}+v_{y}^{2}}
=9α2t2+9β2t4=\sqrt{9 \alpha^{2} t^{2}+9 \beta^{2} t^{4}}
=3t2=α2+β2=3 t^{2}=\sqrt{\alpha^{2}+\beta^{2}}