7.5 Motion graphs

Question 1

Explain the difference between distance and displacement using a ball thrown in the air and caught when it returns as an example.

Answer 1

If the ball rises to a maximum height of 2m, on returning to the thrower tis displacement from its initial position is 0.
However the distance it has travelled is 4m.

Question 2

Describe the displacement-time graph of a ball thrown into the air and caught again.

Answer 2

• immediately after leaving the throwers hand, the velocity is positive and large so the gradient is positive and large.
• as the ball rises, its velocity decreases so the gradient decreases
• at maximum height, its velocity is 0 so the gradient is 0
• as the ball descends, its velocity becomes increasingly negative, corresponding to increasing speed in a downward direction.
• So, the gradient becomes increasingly negative

Question 3

Describe the graph of a ball being thrown up and caught again on a velocity-time graph

Answer 3

(straight diagonal line with negative correlation)
1. the gradient of the line is constant and negative (equal to the acceleration of free fall)

(the acceleration of the object is the same as it descends as to ascending so the gradient of the line is always equal to 9.8m/s^2)

2. the area under the line represents the displacement of the object from its starting position.
   (area in positive section of line and time axis= displacement during ascent)
   (area under negative section of line and time axis= displacement during descent.)