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Question

Physics Question on Motion in a plane

The co-ordinates of a moving particle at any time t't' are given by x=αt3x = \alpha t^3 and y=βt3y = \beta t^3. The speed to the particle at time t't' is given by

A

3tα2+β23t\sqrt{\alpha^{2}+\beta^{2}}

B

3t2α2+β23t^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+β23t^2\sqrt{\alpha^{2}+\beta^{2}}

Explanation

Solution

Speed =v=(3αt2)2+(3βt2)2=3t2α2+β2.=\left|\vec{v}\right|=\sqrt{\left(3\alpha t^{2}\right)^{2}+\left(3\beta t^{2}\right)^{2}}=3t^{2}\sqrt{\alpha^{2}+\beta^{2}}.