Question:

The velocity-time graph below represents the motion of an object. Refer to the graph to answer the following questions.

1. At which time interval is the object moving with a constant velocity? Justify your answer.

2. Determine the total displacement of the object during the time interval from t = 0 s to t = 7 s.

Answer:

1. The object is moving with a constant velocity during the time interval from t = 3 s to t = 5 s.

To justify this answer, we observe that the velocity-time graph is a horizontal line from t = 3 s to t = 5 s, indicating that the velocity remains constant during this time interval. Since velocity is the rate of change of displacement, a constant velocity indicates that the object is moving with a constant displacement per unit time, exhibiting uniform motion.

2. To determine the total displacement of the object during the time interval from t = 0 s to t = 7 s, we need to find the area enclosed by the velocity-time graph during this interval.

The velocity graph consists of two rectangular regions and a triangular region. We will find the area of each region separately and sum them up.

a) Rectangular Region 1:

The width of this rectangle is (5 - 3) s = 2 s, and the height is 5 m/s. Therefore, the area of this region is given by:

Area₁ = width × height = 2 s × 5 m/s = 10 m

b) Rectangular Region 2:

The width of this rectangle is (7 - 5) s = 2 s, and the height is -5 m/s. Since the velocity is negative, the displacement in this region will also be negative. Therefore, the height should be considered as -5 m/s. The area of this region is given by:

Area₂ = width × height = 2 s × (-5 m/s) = -10 m

c) Triangular Region:

The base of the triangle is (7 - 0) s = 7 s, and the height is -5 m/s. We need to consider the absolute value of the height since the displacement is always positive. Therefore, the height should be considered as 5 m/s. The area of this triangular region is given by:

Area₃ = (1/2) × base × height = (1/2) × 7 s × 5 m/s = 17.5 m

Now, we can determine the total displacement (net area) during the time interval from t = 0 s to t = 7 s by summing up the areas:

Total Displacement = Area₁ + Area₂ + Area₃ = 10 m + (-10 m) + 17.5 m = 17.5 m

Therefore, the total displacement of the object during the time interval from t = 0 s to t = 7 s is 17.5 meters.