#322 Two forces, vector F1 = (5i - 6j) N and vector

Two forces, vector F1 = (5i - 6j) N and vector - Physics

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Free Chegg Question

Two forces, vector F1 = (5i - 6j) N and vector F2 = (2i- 3j) N, act on a particle of mass 2.20 kg that is initially at rest at coordinates (+1.55 m, -3.65 m).
(a) What are the components of the particle's velocity at t = 10.3 s?
vector v = _____ m/s
(b) In what direction is the particle moving at t = 10.3 s?
_______° ( counterclockwise from the positive x axis)
(c) What displacement does the particle undergo during the first 10.3 s?
?vector r = _______ m
(d) What are the coordinates of the particle at t = 10.3 s?
x = ____ m
y = ____ m

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Free Chegg Answer

Two forces, ₁ = (5  - 6 ) N and ₂ = (2  - 3 ) N, act on a particle of mass 2.20 kg that is initiallyat rest at coordinates (1.55 m, -3.65 m).
(a) What are the components of the particle's velocityat t = 10.3 s?

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using the Newton second law we can find theacceleration a = f / m so m = 2.20kg then F = F1 + F2;
F = ( 7i - 9j ) so the acceleration a = ( 3.18i - 4.09 j ) m / s^2.

we know that v(t) = a*t

so the particle's velocity at t = 10.3 s is V = (32.754 i - 42.127 j ) m /s.

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> (b) In what direction is the particle moving att = 10.3 s?

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> The direction is given by α = arc tan (Vj /Vi) = arctan(-42.127 / 32.754) = -52.13 degrees = (-52.13 + 180) = 127.87 degrees in the positive direction.

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>(c) What displacement does the particle undergo duringthe first 10.3s?

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> it is given by X = .5a*t^2, so
> we've got X = (168.68i - 216.95j) m.
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> (d) What are the coordinates of the particle at t = 10.3 s?

**
> the initials coordinates are x = ( 1.55 m, - 3.65m ) so the coordinates of the particle at t = 10.3 s is
> Xf = X + x = ( 170.23 i - 220.6 j ) m.