Question

Two forces, vector F 1 = (−6.95î − 2.25ĵ) N and vector F 2 = (−4.25î − 7.05ĵ) N, act on a particle of mass 2.30 kg that is initially at rest at coordinates (+1.95 m, −4.20 m). (a) What are the components of the particle's velocity at t = 10.7 s? vector v = m/s (b) In what direction is the particle moving at t = 10.7 s? ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 10.7 s? Δvector r = m (d) What are the coordinates of the particle at t = 10.7 s? x = m y = m

full explanation please

Answer 1

acceleration = F/m

F = F₁ + F₂

F = (−6.95î − 2.25ĵ) + (−4.25î − 7.05ĵ)

F = - 11.2 i - 9.3 j

a = F/m

a = - 4.87 i - 4.04 j

initial velocity u = 0

velocity at time t is given by

v = u + at

v = 0 + (- 4.87 i - 4.04 j)10.7

v = -52.11 i - 43.23 j m/s

direction is given by

since it is third quadrant direction angle counterclockwise from the +x-axis is 180 +39.7 = 219.7 ^o = 220⁰ approx

c) displacement is given by

x = ut + (1/2)at²

x = 0 + (1/2)(- 4.87 i - 4.04 j)(10.7)²

x = - 278.8 - 231.3 j

displacement is

x = 362.3 m

d) x₂ - x₁ = -278.8

x₂ = x₁ -278.8 = +1.95 -278.8 = -276.8

y₂ - y₁ = -231.3

_y2 = y₁ -278.8 = -4.20 - 231.3 = -235.5

the coordinates are ( -276.8 , -235.5)