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 acceleration, we can use the formula:
a = (v - u) / t
where:
a = acceleration
v = final velocity = 8i - 20j m/s
u = initial velocity = 2i + 4j m/s
t = time interval = 3 sec
Plugging in the values:
a = ((8i - 20j) - (2i + 4j)) / 3
a = (6i - 24j) / 3
a = 2i - 8j m/s^2
Therefore, the acceleration of the particle is 2i - 8j m/s^2.
B) To find position, we can integrate the acceleration function twice.
From the acceleration function, we have:
a = 2i - 8j m/s^2
Integrating with respect to time, we get the velocity function:
v = ∫(2i - 8j) dt
v = 2it - 8jt + C1
Given that the initial velocity is (2i, 4j), we can find the value of C1:
2i = 2(0) - 8(0) + C1
C1 = 2i
Therefore, the velocity function is:
v = 2it - 8jt + 2i
Next, integrating the velocity function with respect to time, we get the position function:
r = ∫(2it - 8jt + 2i) dt
r = it^2 - 4jt^2 + 2it + C2
Given that the initial position is (0,0), we can find the value of C2:
0 = 0 - 0 + 0 + C2
C2 = 0
Therefore, the position function is:
r = it^2 - 4jt^2 + 2it
Now, plugging in the time interval (3 sec), we get:
r = 3(3)^2 - 4(3)^2 + 2(3) + 0
r = 27i - 36j + 6i
r = 33i - 36j
Therefore, the position of the particle at the end of the time interval is 33i - 36j.