The position of particle is given by r=3.0ti+2.0t^2+4.0k . Find the magnitude and direction of velocity of the particle at t=2.0s.
Answers
Answer:
v(t) = (3.0 i - 4.0t j)
a = -4.0 j
The position of the particle is given by:
r = 3.0t i -2.0t2 j +4.0 k
Velocity v of the particle is given as:
v = dr / dt = d (3.0t i -2.0t2 j + 4.0 k) / dt
∴ v = 3.0 i -4.0t j
Acceleration a of the particle is given by:
v = (32 + (-8)2)1/2 = (73)1/2 = 8.54 m/s
Direction, θ = tan-1(vy/vx)
= tan-1(-8/3) = -tan-1(2.667)
= -69.450
The negative sign indicates that the direction of velocity is below the x-axis
v(t) = (3.0 i - 4.0t j)
a = -4.0 j
The position of the particle is given by:
r = 3.0t i -2.0t2 j +4.0 k
Velocity v of the particle is given as:
v = dr / dt = d (3.0t i -2.0t2 j + 4.0 k) / dt
∴ v = 3.0 i -4.0t j
Acceleration a of the particle is given by:
v = (32 + (-8)2)1/2 = (73)1/2 = 8.54 m/s
Direction, θ = tan-1(vy/vx)
= tan-1(-8/3) = -tan-1(2.667)
= -69.450
The negative sign indicates that the direction of velocity is below the x-axis.