a particle is moving in a straight line the particle starts with speed 5ms and accelerates at a constant rate of 2ms for 8s it then decelerates at a constant rate coming to rest in a further 12s
find the total distance during 20s
it is 230 but i don't understand how
5ms+16=21ms initial velocity (io)
the delta v(t)=(ending velocity-io)/total deceleration time = (t)f/s^2
Thus, change in velocity = a t
0-21 = (t)f/s^2
second to top = t
and therefore:
d = Vi t + (1/2) a t^2
To find the total distance traveled by the particle during the given time period, we can break the motion into two separate parts: the acceleration phase and the deceleration phase.
1. Acceleration phase:
The particle starts with an initial speed of 5 m/s and accelerates at a constant rate of 2 m/s² for 8 seconds. To calculate the distance traveled during this phase, we can use the following kinematic equation:
distance = initial speed * time + (1/2) * acceleration * time²
Substituting the given values:
distance = 5 * 8 + (1/2) * 2 * 8²
distance = 40 + 64
distance = 104 meters
2. Deceleration phase:
During the deceleration phase, the particle comes to rest in a further 12 seconds. Since the particle is decelerating, we can treat the deceleration as a negative acceleration. The final speed is zero (rest), so we need to determine the deceleration rate.
Using the formula:
final speed = initial speed + (acceleration * time)
0 = 5 + (deceleration * 12)
We can determine that the deceleration rate is -5/12 m/s².
To find the distance traveled during this phase, we can use the kinematic equation again:
distance = initial speed * time + (1/2) * acceleration * time²
Substituting the given values:
distance = 5 * 12 + (1/2) * (-5/12) * 12²
distance = 60 + (-2.5) * 12
distance = 60 - 30
distance = 30 meters
Now, we can find the total distance traveled by adding the distances covered during the acceleration and deceleration phases:
Total distance = distance during acceleration + distance during deceleration
Total distance = 104 + 30
Total distance = 134 meters
Therefore, the total distance traveled by the particle during the 20-second interval is 134 meters, not 230 meters as you mentioned.