Question

A particle moves along a path y = x2 – 6x + 200. if vx = 6 m/s , what is the value of vy at x = 30?

Answer 1

Your answer is 324 m/s. ☺️

Answer 2

The value of vy = 324 m/s at x = 30 m.

A particle moves along a path y = x² – 6x + 200. if vₓ = 6 m/s.

We have to find the value of vy at x = 30 m.

y = x² - 6x + 200

differentiating both sides with respect to time, t,

dy/dt = 2x dx/dt - 6 dx/dt

We know, " the rate of change of distance with respect to time is known as speed."

$so,\frac{dy}{dt}=v_y\\\\\frac{dx}{dt}=v_x$

$\therefore v_y=2xv_x-6v_x$

Here given,

vₓ = 6 m/s , x = 30 m

$\implies v_y=2(30)6-6(6)=360-36=324m/s$

Therefore the vertical component speed of particle is 324 m/s.