A particle moving in x--y plane has initial velocity (u) of ( 2i , 4j ) m/s and final velocity (v) of (8i , -20j ) m/s in time interval of 3 sec.
Find
A) acceleration
B) position
A) To find the acceleration, we can use the formula:
acceleration = (change in velocity) / (time interval)
The change in velocity (Δv) can be found by subtracting the initial velocity from the final velocity:
Δv = v - u = (8i , -20j) - (2i , 4j) = (6i , -24j) m/s
Now, we can calculate the acceleration using the formula:
acceleration = Δv / time interval = (6i , -24j) / 3 = (2i , -8j) m/s^2
Therefore, the acceleration of the particle is 2i - 8j m/s^2.
B) To find the position, we need to integrate the velocity with respect to time.
Given u = (2i , 4j) m/s, we can integrate each component separately:
x = ∫(2i) dt = 2∫dt = 2t + C1
y = ∫(4j) dt = 4∫dt = 4t + C2
Using the initial conditions at t = 0, we know that x = 0 and y = 0, so the constant terms cancel out:
x = 2t
y = 4t
Therefore, the position of the particle is given by the equations x = 2t and y = 4t.