A particle is initially at the origin. The particle has a constant acceleration of (2i - 3j).?

Question

2 seconds after it has left the origin the velocity of the particle is (2i+14j)ms. The unit vectors i and j are perpendicular.

a)Show that the initial velocity of the particle is(-2i+20j)ms?

b) Find the distance of the particle from the origin after 3 seconds?

c)The speed of the particle is 10ms for the first time T seconds after it has left the origin. Find T?

Answer 1

a)
u = ?, v = 2i+14j, a = 2i-3j t = 2
v = u + at
2i+14j = u + (2i-3j)(2)
2i+14j = u + 4i-6j
2i-4i+14j+6j = u
u = -2i+20j (QED)

b)
u = -2i+20j, s = ?, t = 3, a = 2i-3j
s = ut + (1/2)at^2
s = (-2i+20j)(3) + (1/2)(2i-3j)(3^2)
s = -6i+60j + (1/2)(18i-27j)
s = -6i+50j + 9i-13.5j
s = 3i+36.5j

c)
u = -2i+20j, v = 10, t = T, a = 2i-3j
v = u + at
v = -2i+20j + (2i-3j)(T)
v = (2T-2)i + (20-3T)j

As the magnitude of v is 10,
10 = sqrt[(2T-2)^2 + (20-3T)^2]
10^2 = 4T^2 - 8T + 4 + 400 - 120T + 9T^2
100 = 13T^2 - 128T + 404
13T^2 - 128T + 304 = 0
T = {128+-sqrt[(128^2)-(4.13.304)}/(2.13)
T = {128+-24}/ 26
T = 5.85s and 4s

The speed of the particle would be at 10m/s at 4s and 5.85s.

Answer 2

a) Vintial = Vfinal - (A*t)
(-2i+20j) = (2i+14j) - (4i-6j) = (-2i+20j)
b)distance = Vinital * t + 1/2 * acceleration *t²
distance = (-2i +14j) *3 + 1/2 * (2i-3j) * 9
distance = (-6i + 42j) + (18i - 13.5j) = (12i +28.5j)
c) im bored

Answer 3

It's Friday for Gods sake man. Go out and have a few beers.

Answer 4

I will advise you go open your books and study hard, instead of throwing your tutorial questions here.

Your teacher will be disappointed in you