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 :

The magnitude and direction of the instantaneous velocity of the particle at t=0.25 s is 1.5624 m/s and 36.87°.

What is instantaneous velocity?

The time rate of change of velocity at a particular instant of time is called the instantaneous velocity.

Given the position vector of particle moving in xy plane,  x =2.5m/s²)t²  and y=(5.0m/s³)t³

The instantaneous velocity is v = dx/dt or dy/dt

Instantaneous velocity in x direction v(x) =d/dt(2.5t²) = 5t

At 0.25 sec, Instantaneous velocity in x direction v(x) = 5 x 0.25 = 1.25 m/s

Instantaneous velocity in y direction v(y) =d/dt(5t³) = 15t²

At 0.25 sec, Instantaneous velocity in x direction v(y) = 15 x (0.25)² = 0.9375 m/s

The magnitude of velocity v = √[(v(x)² + v (y)²] = 1.5624 m/s

The direction of velocity is

tan θ= v(y)/v(x)

tan θ=  15t²/5t = 3t

At 0.25 s, tan θ =3 x 0.25 =0.75

θ = tan⁻¹ (0.75) = 36.87°

Thus, magnitude and  direction of the instantaneous velocity of the particle at t=0.25 s is 1.5624 m/s and 36.87°.

Learn more about instantaneous velocity.

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