Question

A particle moves in the xy plane with a constant acceleration given by a = -4.0j m/s^2. At t = 0, its position and velocity are 10i m and (-2.0i +8.0j) m/s, respectively. What is the distance from the origin to the particle at t = 2.0 s?

Answer 1

a(t) = 0i - 4.0j

v(t) = ci - 4t+dj
v(0) = -2i+8j so c=-2 and d=8
so, v(t) = -2i + (8-4t)j

s(t) = (-2t+c)i + (8t-2t^2+d)j
s(0) = 10i, so c=10 and d=0
s(t) = (10-2t)i + (8t-2t^2)j
|s(2)| = |6i+8j| = 10