Question

Using the graph below: What was the instantaneous velocity at t = 12 seconds? What was the distance and displacement of this object between 0 and 15 seconds? What was the average speed and average velocity of this object between 0 and 15 seconds? Finally, what would the velocity-time graph be for this motion?

Answer 1

Answer:

1) 2.5 m/s

2) The distance is 70 m

The displacement is 40 m

3) The average speed is 4.67 m/s

The average velocity is 2.67 m/s

4) The velocity time graph will give the acceleration of the motion

Explanation:

1) The instantaneous velocity is given by the slope at 12 seconds as follows;

Since the distance varies directly with the time from 11 to 15 seconds, the velocity is constant within that time period and is given by the relation;

Velocity = Displacement/(Time duration of displacement)

Velocity = (Final position - Initial position)/(Time duration for the change in position)

Velocity = (-10 - (-20))/(15 - 11) = (-10 + 20)/4 = 10/4 = 2.5 m/s

2) The distance moved by the object between 0 and 15 seconds is given as follows;

Distance = Position 30 to position -25 and from position -25 to position -10

Distance = 30 - (-25) + (-10 - (-25)) = 55 + 15 = 70 m

The displacement = The overall change in position of the object which is the difference between the first and last position;

Displacement = 30 - (-10) = 30 + 10 = 40 m

3) The average speed = Distance/Time = 70/15 = 14/3 = 4.67 m/s

The average velocity = Displacement/Time = 40/15 = 2.67 m/s

4) The velocity time graph will indicate the acceleration of the motion.