A particle p is initially at the point (2, 6) in relation to an origin O, and is moving with velocity (3i + j) m/s. It has a constant acceleration (16i + 24j)m/s^2. Show that after 2 seconds it is moving directly away from O and find its speed at that time
u = x component of speed = 3 + 16 t
v = y component of speed = 1 + 24 t
at t = 2
u = 3 + 32 = 35
v = 1 + 48 = 49
slope of path at t = 2 = v/u = 49/35
location at t = 2:
x = 2 + 3(2) + (1/2)(16)(4) = 40
y = 6 + 1(2) +(1/2)(24)(4) = 56
Tangent to path at t = 2
y = (49/35) x + b
56 = (49/35)40 + b
56 = 56 + b
b = 0
so
y = 1.4 x which goes through the origin
oh, speed at t = 2
we got u = 35 and v = 49
speed = sqrt (35^2 + 49^2)
To determine if the particle is moving directly away from the origin O after 2 seconds, we need to find its position after 2 seconds. This can be done using the equations of motion.
Step 1: Find the displacement after 2 seconds
First, calculate the displacement of the particle after 2 seconds using the equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 3i + j m/s
Acceleration (a) = 16i + 24j m/s^2
Time (t) = 2 seconds
Plugging in the values:
displacement = (3i + j) * 2 + (1/2) * (16i + 24j) * (2^2)
displacement = (6i + 2j) + (1/2) * (16i + 24j) * 4
displacement = 6i + 2j + (8i + 12j) * 2
displacement = 6i + 2j + 16i + 24j
displacement = 22i + 26j
After 2 seconds, the displacement of the particle with respect to the origin O is (22i + 26j).
Step 2: Determine if the particle is moving directly away from O
To determine if the particle is moving directly away from O, we need to check if the displacement is parallel to the initial position vector (2i + 6j).
The direction of the displacement vector is given by the ratio of its components to the initial position vector:
ratio_i = displacement_i / position_i
ratio_j = displacement_j / position_j
Given:
displacement = 22i + 26j
position = 2i + 6j
Calculating the ratios:
ratio_i = 22 / 2 = 11
ratio_j = 26 / 6 ≈ 4.33
Since the ratios of components are different, the displacement vector is not parallel to the initial position vector. Hence, the particle is not moving directly away from O after 2 seconds.
Step 3: Find the speed of the particle after 2 seconds
To find the speed (magnitude of velocity) of the particle after 2 seconds, we need to calculate the magnitude of the velocity vector.
The velocity (v) is given by:
velocity = initial velocity + acceleration * time
Given:
Initial velocity (u) = 3i + j m/s
Acceleration (a) = 16i + 24j m/s^2
Time (t) = 2 seconds
Plugging in the values:
velocity = (3i + j) + (16i + 24j) * 2
velocity = 3i + j + 32i + 48j
velocity = 35i + 49j
The speed of the particle after 2 seconds is the magnitude of the velocity vector:
speed = |35i + 49j| = sqrt((35^2) + (49^2))
Calculating the speed:
speed = sqrt(1225 + 2401)
speed = sqrt(3626)
Therefore, the speed of the particle after 2 seconds is approximately 60.21 m/s.