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→