if the position vector of a particle is given by r=3t2 i +5t j + 4k find the (a) the velocity of the particle at t=3s
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We have the position vector given in terms of time t. r(t) = t^3*i + t^2*j
To find the velocity vector we have to differentiate r(t) with respect to time.
r'(t) = 3t^2*i + 2t*j
The vector representing acceleration is the derivative of the position vector
r''(t) = 6t*i + 2*j
When time t = 2.
The velocity vector is 3*2^2*i + 2*2*j
=> 12*i + 4*j
The speed is the absolute value of the velocity vector or sqrt(12^2 + 4^2) = sqrt (144 + 16) = sqrt 160
The acceleration vector is 6*2*i + 2*j
=> 12*i + 2*j
The required acceleration at t=2 is 12*i + 2*j and the speed is sqrt 160
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