Can anyone help me in solving this numerical The position of a particle moving along X Axis is given by; x= At + 4 Btcube What is its velocity at t=5s

Question

Answer

v(t) = d( A*t + 4B*t³)/DXv(t) = A + 4B*(3t²)v(5)= A + 4B*3*5²v(5)= A + 300B

Answer

A +300B

Answer

A + 140B unit/s

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A+300B

Answer

A

Answer

find d÷dx apply value of x in that then divide by time

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apply d/dt to the equation     sub 5 in t for dx/dt you will get       A plus 300B

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velocity of particle is A m/s

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A +4

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simple..diffreciate nd put t=5 sec

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A+300B

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prajil

Answer

v=A+12Bt^2

Answer

u mistake aishwarya

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