# The position vector of a particle is given by r(t)=t^3*i+t^2*j. What are it's velocity speed and acceleration when t=2

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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**.

The velocity is given by the 1st derivative of r(t).

We'll differentiate r(t) with respect to t.

v(t) = r'(t) = 3t^2*i+ 2t*j

The acceleration is given by the 1st derivative of v(t).

We'll differentiate v(t) with respect to t.

a(t) = v'(t) = 6t*i + 2j

The speed is:

|v(t)| = sqrt[(3t^2)^2 + (2t)^2]

|v(t)| = sqrt(9t^4 + 4t^2)

We'll put t = 2 and we'll get:

v(2) = 3*(2)^2*i+ 4*j

v(2) = 12i + 4j

a(2) = 12i + 2j

|v(2)| = sqrt(144+ 16)

|v(2)| = sqrt(160)

|v(2)| = 4sqrt10

**The velocity and acceleration of the particle, when t=2, are: v(2) = 12i + 4j ; a(2) = 12i + 2j and |v(2)| = 4sqrt10.**