Find its position at t=1.

Question

A particle starts at the origin with initial velocity i+j-k. Its acceleration is a(t)=ti+j+tk. Find its position at t=1.

Answer 1

so if acceleration is a(t) = ti + j + tk

velocity is the antiderivative of acceleration, so we just do antiderivative of each term of acceleration

a(t) = ti+j+tk

v(t) = (1/2t^2+C1)i + (t+C2)j + (1/2t^2+C3)k, where C1, C2, and C3 are the three components of initial velocity (which they give us)

so v(t) = ( 1/2t^2 + 1 )i + ( t + 1 )j + ( 1/2t^2 - 1)k

Then for position, we just do antiderivative again

s(t) = ( 1/6t^3 + t + C4)i + (1/2t^2 + C5)j + (1/6t^3 - t + C6)k, where C4, C5, and C6 are the three components of the initial position (which are all 0 since we start at the origin)

so s(t) = (1/6t^3 + t)i + (1/2t^2)j + (1/6t^3 - t)k

now to find position at t = 1, we just plug in 1

so s(1) = (1/6 + 1)i + (1/2)j + (1/6 - 1)k = (7/6)i + (1/2)j - (5/6)k