Question

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The

object moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the
displacement of the object?
O the area of the rectangle under the line
the area of the rectangle above the line
the area of the rectangle plus the area of the triangle under the line
the area of the rectangle plus the area of the triangle above the line
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Answer 1

Answer: The correct answer is:

the area of the trapezoid under the line

To explain this, let's consider the velocity vs. time graph again. Since the object moves with constant acceleration, the graph will be a straight line with a positive slope. The area under the line represents the distance traveled by the object, which is equal to the displacement if the initial position is zero.

The area under the line is a trapezoid because the velocity is changing over time. The base of the trapezoid is the time interval, and the heights are the initial and final velocities. The formula for the area of a trapezoid is:

Area = (base1 + base2) / 2 * height

where:

base1 = initial velocity

base2 = final velocity

height = time interval

Substituting the given values, we get:

Area = (v_i + v_f) / 2 * t

where v_i = 3 m/s, v_f = 10 m/s, and t is the time interval over which the velocities change.

Therefore, the correct answer is the area of the trapezoid under the line.