Answer: a) vx = 32.8 m/s vy= -42.1 m/s
b) - 52º from x axis.
Explanation:
a) as both forces are external to the particle and are acting simultaneously, we can find the total force Ft, as the vector sum of F1 and F2, as follows:
F = F₁ + F₂ = 7i -9j N
We can apply then the Newton's 2nd Law, as follows:
F = m. a
This vector equation, can be decomposed in two independent algebraic equations, one for each axis:
Fx = m. aₓ
Fy = m. ay
As we know Fx, Fy and m, we can get ax and ay, as follows:
aₓ = 7 N/ 2.20 Kg = 3.2 m/s²
ay = -9 N /2.20 Kg = -4.1 m/s²
By definition, we know that a = vf-v₀ / t (assuming that t₀ = 0).
Now, as the particle starts from rest, this means that v₀ = 0.
So, we have a vector equation, that we can solve for the components of v, at any time, in this case, at t = 10.3 s.
Solving for vₓ and vy, we have the following:
vₓ = aₓ. t = 3.2 m/s² . 10.3 s = 32.8 m/s
vy = ay. t = -4.1 m/s² . 10.3 s = -42.1 m/s
b) In order to find the direction in which the particle is moving, we need to find simply the quotient between vy and vx, as this quotient is equal to the tangent that the velocity vector does with the x axis, as follows:
tg θ = vy /vx = -42.1 / 32. 8 = -1.29
θ = arctan (-1.29) = -52º
