The position of a particle is given by
Where t is in seconds and the coefficients have the proper units for r to be in metres.
A. Find the v and a of the particle?
B. What is the magnitude and direction of velocity of the particle at t = 2.0 s?
Given:
Position of the particle =
A. velocity is rate of change of displacement w.r.t time
Velocity = v= dr/dt
= 3ti-4j m/s
Acceleration = a = dv/dt
= 3i m/s2
B. At t=2.0s
Velocity = v = 3ti-4j m/s
= 3(2)i-4j m/s
= 6i - 4j m/s
Magnitude of velocity =
= 7.211
tanθ = -4/6 =0.667
θ = -33.7˚
Acceleration = a = 3i m/s2
Magnitude of acceleration = 3
tanθ = 0
θ = 0˚
Velocity is 7.211 m/s at an angle of -33.7˚ with the horizontal. Acceleration is 3 m/s2 in the X direction.