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    Graphing with time, velocity, and acceleration

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    The velocity of a car on a straight stretch of highway is recorded every 5 seconds. The data is in the table below. A negative velocity means that the car is traveling east. A positive velocity means it is traveling west. Make a velocity versus times graph, that is, plot velocity on the vertical axis and time on the horizontal axis. Base your answers to the questions below on the data and your graph.

    Time (s) Velocity (m/s)
    0 -32
    5 -24
    10 -16
    15 -8
    20 0
    25 8
    30 16
    35 24
    40 32
    45 40
    50 48

    Now, I can sketch the graph and answer most of my question except for nine questions I need your help!

    1) What is the value of the slope of the velocity versus time graph, in units of m/s^2?

    a 8/25, b 5/8, c 4/5, d 8/5, e none of above

    2) The slope of a velocity versus time graph is equal to the derivative of the velocity with respect to time, which is called acceleration. What is the acceleration of the car at time = 20 seconds, in units of meters/second^2?

    a zero, b 5/8, c 4/5, d 8/5, e none of above

    3) For the time from t=0 to t=50 seconds, the value of acceleration of the car (in units of m/s^2)...

    a is zero, b increases, c goes from a negative value to a positive value, d decreases, then increases, e is constant

    4) The change in position of an object, also known as its displacement, can be calculated by multiplying its average velocity times the time. If velocity is measured in units of m/s, and time is measured in units of seconds, the displacement will be i units of....

    a meters, b m/s^2, c seconds/m^2,
    d (m/s)/(seconds), e none of above

    5) The change in position of an object (displacement) can be calculated from a velocity versus time graph by calculating the area between the velocity graph and the horizontal axis. Area above the "v=0" axis is positive, area below the "v=0" axis is negative. What is the displacement of the car between the time t=0 and t=20 seconds?

    a 640 m, b 320 m, c -320 m, d -640 m, e none of above

    6) What was the displacement of the car from time t =20 seconds to t=50 seconds, in units of meters?

    a 1440, b 1200, c 720, d 8/5, e none of above

    7) At time t=50 seconds, where is the car located compared to its location at t=0 seconds?

    a 1040 meters away
    b 400 meters east
    c 400 meters west
    d 800 meters east
    e none of above

    8) Graphs can provide lots of information. In a velocity versus time graph, you can (A) read off the graph the velocity of the object at any time on the graph, (B) find the acceleration of the object at any time by calculating the slope of the graph at that time (C) find the displacement from one time to another time by calculating the area under the graph. Check all the statements that can be made about the motion of the car at t=15 seconds?

    a the car's velocity at t=15 sec. is -8 m/s.
    b The car's acceleration at t=15 sec. is (8/5) m/s^2.
    c The car traveled 60 meters east since t=10 seconds.
    d The cars acceleration is changing.

    9) Which description of motion of the car, from t=0 to t=50 seconds is correct, accurate and unambiguous? (That is, it correctly and accurately describes the motion of the car, without allowing another interpretation.)

    a The car slows down to zero, and then speeds up.
    b The car's acceleration increases at a constant rate.
    c The car's velocity is constant.
    d The cars' displacement changes at a constant rate.
    e The car's velocity changes at a constant rate, and changes direction from east to west.

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    Solution Summary

    This solution looks at a kinematics problem: time, velocity, acceleration.