Question

A particle's trajectory is described by x =(1/2t3?2t2)m and y =(1/2t2?2t)m, where t is in s, What is the particle's speed at t=4.5s ?

Answer 1

As we know that it is given to us

$x = (\frac{1}{2}t^3 - 2t^2)$

$y = (\frac{1}{2}t^2 - 2t)$

now we know that rate of change in position is known as velocity

so here we will have

$v_x = \frac{dx}{dt}$

$v_x = \frac{3}{2}t^2 - 4t$

similarly

$v_y = \frac{dy}{dt}$

$v_y = t - 2$

now we have t = 4.5 s

$v_x = 1.5(4.5)^2 - 4(4.5) = 12.375 m/s$

$v_y = 4.5 - 2 = 2.5 m/s$

now the net speed is given as

$v^2 = v_x^2 + v_y^2$

$v^2 = 12.375^2 + 2.5^2$

$v = 12.625 m/s$