A particle accelerates uniformly from rest at 6.0ms for 8s and then decelerate uniformly to rest in the next 5s . determine the magnitude of the deceleration
oh, come on.
v = at, so
a = v/t = -48/5 = -9.6 m/s^2
Apologies for the mistake. You are correct.
Using equation 1: v = u + at
For the deceleration phase, the initial velocity (u) is 48 m/s (final velocity of the acceleration phase) and the time (t) is 5 seconds (duration of the deceleration phase). The final velocity (v) is 0.
0 = 48 + a(5)
Rearranging the equation:
48 = 5a
a = 48/5 = -9.6 m/s^2
Therefore, the magnitude of the deceleration is 9.6 m/s^2.
To solve this problem, we will use the equations of motion for uniformly accelerated motion:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
In the first 8 seconds, the particle accelerates uniformly from rest (u = 0) at 6.0 m/s^2. We need to find the final velocity after 8 seconds.
Using equation 1, we have:
v = u + at
v = 0 + (6.0)(8)
v = 48 m/s
In the next 5 seconds, the particle decelerates uniformly to rest (v = 0). We need to find the magnitude of the deceleration.
Since the final velocity is 0 and the initial velocity is 48 m/s, we can use equation 3 to solve for the magnitude of deceleration (a):
v^2 = u^2 + 2as
(0)^2 = (48)^2 + 2a(5)
0 = 2304 + 10a
10a = -2304
a = -230.4 m/s^2
Therefore, the magnitude of the deceleration is 230.4 m/s^2.