# Physics-calc

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Suppose the position of an object is given by ->r(vector) = (3.0t^2*ihat - 6.0t^3*jhat)m. Where t in seconds.

Determine its velocity ->v as a function of time t.

Determine its acceleration ->a as a function of time t.

Determine ->r at time t = 2.5 s.

Determine ->v at time t = 2.5s.

• Physics-calc -

This is a problem of differentiation where the independent variable is t, and
r is the position vector.
v is the velocity vector.
a is the acceleration vector.

The motion is described in two orthogonal directions i and j, which means that you can do the calculations in each of the directions independently of each other.

Given
r = (3.0t^2i - 6.0t^3j) m
The i and j components of the position vector are
Pi(t)=3.0t^2
Pj(t)=6.0t^3

So d(Pi(t))/dt = d(3t²)/dt = 6t
...
can you complete the rest?

• Physics-calc -

What it P representing? Position?

• Physics-calc -

6t-18t

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