A 1.57-kg particle initially at rest and at the origin of an x-y coordinate system is subjected to a time-dependent force of

Question

F(t) = (7.00ti − 8.00j)N with t in seconds.
(a) At what time t will the particle's speed be 19.0 m/s?

s

(b) How far from the origin will the particle be when its velocity is 19.0 m/s?

m

(c) What is the particle's total displacement at this time? (Express your answer in vector form. Do not include units in your answer.)

The answer keyare: a)=2.51s, b)=19.9m, and c)=11.8i-16.I got my acceleration by dividing N with Kg=(4.46ti+5.10tj)m/s^2 but got stuck here.

Answer 1

(a) you want

(7.00t)^2 + 8.00^2 = 19^2

(b) now plug that t value and integrate twice of the F/m vector (acceleration) with v(0)=s(0)=0

(c) use the distance formula again for the new position

Answer 2

ehh Steve your method doesn't get me anywhere closer to the answer key. Cau you explain futher with steps?

Answer 3

Hmmm. I see I didn't read carefully.
F(t) = <7t,-8>
since F=ma,
a(t) = F/1.57 = <4.458t,-5.096>
so,
v(t) = ∫a(t) dt = <2.279t^2,-5.096t>
|v| = 19 when
(2.279t^2)^2 + (5.096t)^2 = 19^2
t = 2.49

Not sure how they got 2.51, but maybe some roundoff is involved.

Now integrate v to get s(t) and work from that.