Question

2. (a) Explain the difference between constant velocity and constant acceleration. (b) Are you more likely to be able to walk at a constant velocity or at constant acceleration? (c) For the equation x = Xo + of + at, what does a graph of x versus t look like? (d) What does the graph look like if a -0? Please scetch graphs and give explanations. (7 pts) 2 3. For the following position versus time values, calculate the average velocity for each of the four-time intervals. What is the physical meaning of a negative average velocity? (8 pts) 2.23 1.98 Position (m) Time (s) 1.38 1.63 2.08 1.17 1.652.08 2.65 3.16

Answer 1

Ans 2.

a) Constant velocity: When the change in position of object is the same every second. then the object is moving at constant velocity.

Constant acceleration: When the change in velocity of an object is the same every second, then the acceleration is constant.

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b) You are likely to be able to walk at constant velocity as walking at constant acceleration would require your velocity to be increased every second(that is hard).

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c) The graph of x vs t looks like parabola as shown in the figure below.

Consider an object starts at position x_o = 0, initial velocity _^vo and acceleration a. As the time increases the x position of the object increases more rapidly(x is a quadratic in time.)

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d) The graph of x vs t when a = 0 would look like a straight like as shown in the figure below.

An object starts at x = x_o with initial velocity v_o and the acceleration is zero. We can see from the graph the change in displacement of the object is the same every second. So, the object is moving at constant velocity. The velocity of the object v_o is the slope of the graph.

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3.

$\text{ The average velocity of an object over a time interval is given by} \\ v_{av} = \frac{\Delta x}{\Delta t} \\ \\ \Delta x \text{ = the change in the position of the object.} \\ \Delta t \text{ = time interval }$

$\text{Consider first interval,} \\ \Delta x \text{ = the change in the position of the object. = 1.63 - 1.38 = 0..25 m} \\ \Delta t \text{ = 1.65 - 1.17 = 0.48 s } \\ \text{ The average velocity of over the first interval is } \\ v_{av} = \frac{\Delta x}{\Delta t} = \frac{0.28}{0.45} = 0.521 \ m/s \\ \\$

Similarly we find the average velocity for remaining 3 intervals. They are in the table below.

Average velocity(m/s)

0.521

1.40

-0.263

-0.196

The negative average velocity means the object is moving in negative direction. Consider a car moving towards a traffic light. Choosing positive direction of the motion to be away from the traffic light. So the average velocity of the car is negative.