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The vector components of the position of a particle moving in the xy plane as a function of time are given by

rx=x=(2.5m/s2)t2

ry=y=(5.0m/s3)t3



(a) What is the magnitude and direction of the instantaneous velocity of the particle at t=0.25 s?

Respuesta :

indrawboy

Instantaneous velocity :the derivate of the displacement function

vx = 5t

vy = 15t²

at t = 0.25 s

vx = 1.25

vy = 0.9375

magnitude

[tex]\tt v=\sqrt{v_x^2+v_y^2}=\sqrt{1.25^2+0.9375^2}=1.5625~m/s[/tex]

direction

[tex]\tt \theta=tan^{-1}\dfrac{0.9375}{1.25}=36.87^o[/tex]

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