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Given displacement of a particle, r(t) = 3t'i + 4t’j +2tk. Find a) velocity of the particle at t=1sb) acceleration of the particle at t= 1s. |
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Answer» ong>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 |
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