2
6. A particle starts from a point with a velocity of +6.0
m/s and moves with an acceleration of -20 m/s
Show that after 6 s the particle will be at the starting
point.
Answers
Answer:
Explanation:
At fist it seems that you would need to integrate the signed area under the time , velocity graph.
However it’s a simple diagram:
This shows what the graph must look like if your proposition is true. It’s immediately clear that we don’t need to know the area (distance traveled) because the acceleration is uniform and we have two triangles of opposite area only if velocity = 0m/s at time = 3 seconds. We therefore only need to confirm the gradient a.k.a. acceleration required to bring 6m/s to zero in 3 seconds is 20 m/s^2. Is it?
The delta velocity over 3 seconds is -6m/s so:
acceleration = -6m/s / 3s = 2 m/s^2
So the answer is no, you have 10 times the acceleration required. If your question were altered with either +60 m/s starting velocity or -2 m/s^2 acceleration then the above diagram and calculation suitably modified would show that the particle was back at the starting point.
At fist it seems that you would need to integrate the signed area under the time , velocity graph.
However it’s a simple diagram:
This shows what the graph must look like if your proposition is true. It’s immediately clear that we don’t need to know the area (distance traveled) because the acceleration is uniform and we have two triangles of opposite area only if velocity = 0m/s at time = 3 seconds. We therefore only need to confirm the gradient a.k.a. acceleration required to bring 6m/s to zero in 3 seconds is 20 m/s^2. Is it?
The delta velocity over 3 seconds is -6m/s so:
acceleration = -6m/s / 3s = 2 m/s^2
So the answer is no, you have 10 times the acceleration required. If your question were altered with either +60 m/s starting velocity or -2 m/s^2 acceleration then the above diagram and calculation suitably modified would show that the particle was back at the starting point.