Motion Graph Rapid Fire. What is the velocity from 0-2 s ?

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Presentation transcript:

Motion Graph Rapid Fire

What is the velocity from 0-2 s ?

Answer = 2 m/s

Describe the motion from 2 s to 5 s.

Answer = Still

From 5 s to 8 s the object is: (accel away) (accel towards)

Answer = Accelerating away

What is the final position?

Answer = 6m

This is the p vs t graph for a football player tackling a ball carrier. What is located at 0m?

Answer = The running back (the guy he was trying to tackle)

Describe what happened.

Answer = The player slowly ran towards the ball carrier, then accelerated towards the ball carrier to run at a faster speed, and then tackled him.

What direction is the object moving from t=3s to t=5s?

Answer = Left (negative 2.5 m/s)

What does the straight line on a V vs t graph indicate?

Answer = constant velocity

What is the final position of the object? How do I find it?

Answer = I take the area under the curve to find the answer to be: 9m-4m+3m = 8m

If I were to change this graph to a position vs time graph, what would the slope be from 3s to 5s?

Answer = -2 m/s

What does the sloped line from 2s to 4s represent?

Answer = positive acceleration (1m/s 2 )

What is the acceleration from 7s to 8s?

Answer = -3 m/s 2

How can I find the velocity of the object using an acceleration vs time graph?

Answer = find the area under the curve.

Last question… What is the velocity of the object at the end of this motion graph?

Answer = 9 m/s- 9 m/s + 1 m/s = 1m/s