Question

The position of a particle is given by
r=(3.0ti-2.0t^2j+4.0k)m
Where t is in seconds and the coefficient have the proper units for r to be in meters.
(a). Find the velocity and acceleration of the particle.
(b). What is the magnitude and direction of velocity of the particle at t=4.0s?

Answer 1

Answer:

Explanation:

r⃗ =3.0ti^+2.0t^2 j^+4.0k^

We know velocity is given by

v⃗ =dr/dt

v⃗ =3.0i^+4tj^

Velocity after 2 seconds

v⃗ 2=3i^+8j^

Magnitude of velocity = √(〖(3)〗^2+〖(8)〗^2 )=8.54 ms-1

Direction is given by

θ=tan-1(83)=69.5≈70°θ=tan-1(83)=69.5