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The figure shows the velocity graph of a 73 kg passenger in an elevator Figure

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Part A:

What is the passenger’s weight at t = 1s? Express your answer to two significant figures and include the appropriate units.

Part B:

What is the passenger’s weight at t = 5s? Express your answer to two significant figures and include the appropriate units.

Part C:

What is the passenger’s weight at t = 9s? Express your answer to two significant figures and include the appropriate units.

**Velocity vs Time Graphs:**

Velocity vs time graph is a graphical representation of the motion of an object over a period of time. In this period of time the object is subject to variations in its speed and consequently, fluctuations the speed at which it moves. The object’s acceleration can be determined by measuring its slope and the displacement may be calculated by measuring the area beneath the curve.

Given data: m=73 kg is the mass of the passenger

Part A:To calculate the weight of the person at t = 1 sec. It is necessary to determine the speed of the passenger at the moment of. We will calculate how steep the line is in the initial section.

The forces that affect the person in the elevator include his weight (mg) that is acting downwards in conjunction with his reaction (R) from the floor of the elevator that is vertically upward. Thus, by applying Newton’s second law for motion, we can:

R – mg = ma or R = mg + ma.

Here,

Part B:At the time

t = 5 seconds, the passenger is not accelerating, i.e. a=0 m/s2. Therefore:R – mg = 0 or R = mg or R = 73 x 9.8 or R = 720N (correct to two significant figures)

Part C:Let’s find out the acceleration at which the passenger travels with

the speed of 9 seconds.The final step is we determine what the weight of the passenger.

R – mg = ma or R = m(a+g) = 73(-2 + 9.8) = 570N (correct to two significant figures)