Determine if the acceleration of an object is given by a^→ = i^→ + 2 j^→ + 6tk^→ find out the object's velocity and position functions here given that the initial velocity is v^→ (0) = j^→ - k^→ and initial position is as r^→ (0) = i^→ - 2 j^→ + 3k^→.

Solution

We will first obtain the velocity.  To do this all (well almost all) there is a requirement to do is integrate the acceleration.

v^→ (t) = ∫ a^→ (t) dt

= ∫ i^→ + 2j^→ + 6tk^→ dt

= ti^→ + 2t j^→ + 3t² k^→ + c^→

To totally get the velocity we will need to find out the "constant" of integration. We can make use of the initial velocity to get this.

j^→ - k^→ = v^→ (0) = c^→

After that the velocity of the object is,

v^→ (t) = t i^→ + 2t j^→ + 3t² k^→ + j^→ - k^→

= t i^→ + (2t + 1) j^→ + (3t² -1) k^→

We will find out the position function through integrating the velocity function.

r^→ (t) = ∫ v^→ (t) dt

= ∫ t i^→ + (2t + 1) j^→ + (3t 2 -1) k^→ dt

= 1/2 t² i^→ + (t2 + t) j^→ + (t3 - t) k^→ + c^→

By using the initial position gives us,

 i^→ - 2j^→ + 3k^→ = r^→ (0) = c^→

 thus, the position function is,

r^→ (t) = (1/2 t² + 1) i^→ + (t² + t - 2) j^→ + (t3 - t + 3) k^→