If the motion of a particle is governed by the equation X is equals to 20 cube- 3 square + 30 + 2 find the position velocity and acceleration of the particle at time T is equals to 2 second
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Hey Saatwik,
X(t)=20t^3-3t^2+30t+2
=>v(t)=d(X)/dt=20×3t^2 -3×2t +30
=60t^2 -6t+30
=>v(2)=60(4)-6(2)+30=270-12=258m/s
=>a(t)=dv/dt=120t-6
=>a(2)=240-6=234m/s^2
Hope you got it.
X(t)=20t^3-3t^2+30t+2
=>v(t)=d(X)/dt=20×3t^2 -3×2t +30
=60t^2 -6t+30
=>v(2)=60(4)-6(2)+30=270-12=258m/s
=>a(t)=dv/dt=120t-6
=>a(2)=240-6=234m/s^2
Hope you got it.
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