Question

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?

Answer 1

Instantaneous velocity :the derivate of the displacement function

vx = 5t

vy = 15t²

at t = 0.25 s

vx = 1.25

vy = 0.9375

magnitude

`\tt v=√(v_x^2+v_y^2)=√(1.25^2+0.9375^2)=1.5625~m/s`

direction

`\tt \theta=tan^(-1)(0.9375)/(1.25)=36.87^o`