What are the components of the particle's velocity

[tdusffx]

Two forces, F1 = (-6i - 4j)N and F2 = (-3i, 7j)N, act on a particle of mass 2 kg that is initially at rest at coordinates (-2.00m, +4.00m).

(a) what are the components of the particle's velocity at t = 10s

F1 = (-6i-4j)N
F2 = (-3i + 7j)N

Ff = F1 + F2

Ff = (-9i 3j)

F = ma; -9 = 2a = ai = -4.5m/s^2
F = ma; 3 = 2a = aj = 1.5m/s^2

V = -4.5(10)i + 1.5(10)j

V = 45i + 15j

(b) In what direction is the particle moving at t = 10s

Vector F = sqrt ( (-9)^2 + 3^2) = root(90)

tetha = arctan (3/-9)

(c) what displacement does the particle undergo during the first 10s

Xf = Xi + VixT + 1/2AxT^2

deltaX = 0+ 1/2(9/2)(10)^2

deltaX = 225m

Yf = Yi + ViyT + 1/2AyT^2

delta = 0 + 1/2(3/2)(10)^2 = 75m

Are my answers right?

 

[Dick]

Ok for a), except shouldn't it be -45i+15j. Ok for b). In c) check on signs for your displacement vector. It's moving down in x and up in y.

 

[tdusffx]

oh, ok. So delta Y is -75m?

 

[Dick]

Uh, "down in x, up in y"? Shouldn't delta X be negative? Feel free to disagree.

 

[tdusffx]

oh, ya. my bad. You are right, dick.
V = -45i + 15j

and ya, therefore x is negative.