Question

Given displacement of a particle, r(t) = 3t'i + 4t’j +2tk. Find
a) velocity of the particle at t=1s
b) acceleration of the particle at t= 1s.​

Answer 1

Answer:

Magnitude of the velocity of particle is 8 m/s

Explanation:

Position : R=4sin(2πt)i^+4cos(2πt)j^

Velocity: v=4×2πcos(2πt)i^−4×2πsin(2πt)j^

Acceleration: a=−4(2π)2sin(2πt)i^−4(2π)2cos(2πt)j^=−(2π)2R

∴ acceleration is along −R

Magnitude of velocity: ∣v∣=(4×2πcos(2πt))2+(−4×2πsin(2πt))2=(8π)2sin2(2πt)+cos2(2πt)=8π

Magnitude of acceleration : ∣a∣=∣−(2π)2R∣=4π2∣R∣=16π2=

4(8π)2=Rv2

Rx=x=4sin(2πt)

Ry=y=4cos(2πt)

Rx2+Ry2=x2+y2=42

⇒ the path of the particle is a circle of radius