At t = 0, a particle moving in the xy plane with constant acceleration has a vel

Question

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.80 s, the particle's velocity is v = (8.00 i + 5.00 j) m/s.

(a) Find the acceleration of the particle at any time t. (Use t, i, and j as necessary.)
.......................................................... m/s2
(b) Find its coordinates at any time t.
x = ........................................... i m
y = ............................................ j m

Explanation / Answer

The velocity of the particle at t= 0 is v1 = 3.0i - 2.0j m/s The velocity of the particle at t=2.8s, v2 = 8.0i+5.0j m/s The constant acceleration a = v2 - v1 / t = (8.0i+5.0j ) -(3.0i - 2.0j) / (2.80s) = (5.0i +7.0j) / 2.80s = 1.785i + 2.5j d^2s = dt^2(1.785i + 2.5j) s = 0.8925t^2i + 1.25t^2j+ C 0 = C at t = 2.8s s = 0.8925t^2i + 1.25t^2j+ C = 6.99 i + 9.8j so x = 6.99 m y = 9.8

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