Question

Somebody please shows me how to solve for b) and c)step by step. I found a)=2.27s but stuck there. A 1.49-kg particle initially at rest and at the origin of an x-y coordinate system is subjected to a time-dependent force of F(t) = (4.00ti − 8.00j) Nwith t in seconds.(a) At what time t will the particle's speed be 14.0 m/s? (b) How far from the origin will the particle be when its velocity is 14.0 m/s? (c) What is the particle's total displacement at this time? (Express your answer in vector form. Do not include units in your answer.)

Answer 1

F = ma, so
a(t) = 2.68ti-5.37j
v(t) = 1.34t^2i-5.37tj

(a) (1.34t^2)^2 + (5.37t)^2 = 14.0^2
t = 2.27

(b) s(t) = 0.45t^3i-2.68t^2j
s(2.27) = 5.26i-13.81j
|s| = 14.78

(c) do you mean the total distance traveled? That would be

∫[0,2.27] √((1.34t^2)^2+(5.37t)^2) dt
= ∫[0,2.27] t√(1.80t^2+28.84) dt
= 14.89