A car starting from rest accelerates uniformly are attaing a speed of 80m/s in 30s it maintains this stedy speed for another 30s. It then slow down uniformly until it comes to rest in the next 40s
You have not asked a question. You have only presented some data.
Sketch the velocity time graph and show the uniform acceleration during the fist stage of the journey, and the retardation during the last stage of the journey. How long did the car travel altogether?
To understand the scenario described, we can apply the equations of motion for uniformly accelerated motion. We can break the journey of the car into three different intervals: acceleration, constant speed, and deceleration.
1. Acceleration phase:
The car starts from rest and accelerates uniformly until it reaches a speed of 80 m/s in 30 seconds. We need to find the acceleration during this phase.
We can use the equation:
v = u + at
Where:
u = initial velocity (0 m/s)
v = final velocity (80 m/s)
t = time (30 s)
Solving for acceleration (a):
80 = 0 + a * 30
80 = 30a
a = 80/30
a = 2.67 m/s²
Therefore, the acceleration during the first phase is 2.67 m/s².
2. Constant speed phase:
The car maintains a steady speed of 80 m/s for the next 30 seconds. Since there is no change in velocity, the acceleration during this phase is zero.
3. Deceleration phase:
The car slows down uniformly in the next 40 seconds until it comes to rest. We need to find the acceleration during this phase.
Using the same equation:
v = u + at
Where:
u = initial velocity (80 m/s)
v = final velocity (0 m/s)
t = time (40 s)
Solving for acceleration (a):
0 = 80 + a * 40
a = -80/40
a = -2 m/s² (Negative sign indicates deceleration)
Therefore, the acceleration during the deceleration phase is -2 m/s².
In summary:
1. Acceleration phase: The acceleration is 2.67 m/s².
2. Constant speed phase: The acceleration is 0 m/s².
3. Deceleration phase: The acceleration is -2 m/s².