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

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.