Illustration 2.5: Motion on a Hill or Ramp
Please wait for the animation to completely load.
A putted golf ball travels up a hill and then down again
(position is given in meters and time is given in seconds). Restart. When an object (like a
golf ball) travels up or down an inclined ramp or hill,
its motion is often characterized by constant, nonzero acceleration. If the incline of the hill
is constant, then the motion of the object can also be considered straight-line
motion (or one-dimensional motion). It is convenient to analyze the motion
of the golf ball by defining the +x axis to be parallel to the hill and directed
either upward or downward along the hill as shown in Animation 1.
Here are some characteristics of the motion that you should
convince yourself are true:
- In Animation 1 the +x direction is defined to be down the hill. Therefore, when the ball moves down the hill, it is
moving in the +x direction and thus v_{x} is positive. When the ball moves up the hill, it is moving
in the -x direction and thus v_{x} is negative.
- As the golf ball is traveling up/down the hill, is it slowing down or speeding
up? Well, the answer to this question depends on what you mean by slowing down and speeding up. As the ball rolls up the hill its velocity is negative
(because of how the x axis is defined) and decreasing in
magnitude (a smaller negative number). At the top of the hill, its velocity
is zero, and as it travels down the hill, the ball is speeding up. Therefore, its speed decreases, reaches zero, and
then increases. How can this be if v_{x} is always increasing? Speed is the magnitude of
velocity (and is always a positive number). As the ball travels up the hill, v_{x}
increases from -5 m/s to zero; yet its speed
decreases from 5 m/s to zero. Note that the phrases "speeding up" and
"slowing down" refer to how the speed changes, not necessarily to how the velocity
changes.
- Is the acceleration
of the golf ball increasing, decreasing, or constant? To answer this, look at the slope
of the graph at every instant of time. The slope of the velocity vs. time
graph (velocity in the x direction) is equal to the acceleration (in the x
direction). Does it change or is it the same? Notice that it is constant at all times and is in the positive x direction (as defined by
the coordinates).
- Besides using the graph to calculate acceleration, you can also use
the velocity data from the data table. Since
average acceleration is the change in velocity divided by the time interval, choose any time interval, measure v_{xi} and v_{xf}, and calculate the
average acceleration a_{x avg}. Since the acceleration is
constant, the average and instantaneous accelerations are identical.
- The direction of acceleration can also be found by subtracting
velocity vectors pictorially. Animation 2 shows the
black velocity vectors at t = 0.2 s and t = 1.0 s. To subtract vectors,
drag v_{i} away from
its original position (you can drag the little circle on the arrow's tail) and then drag the red vector -v_{i} into place and use the
tip-to-tail method. The direction of the acceleration is in the same direction as
the change-in-velocity vector. Now, try Animation 3,
which shows the velocity vectors at
t = 1.2 s and t = 2.0 s. Compare
the change-in-velocity vector for each of the two time intervals. You will find that
they are
the same. Since the acceleration is constant, the change in velocity is constant for any
given time interval.
- The area under a v_{x} vs. time graph is always the displacement, Δx.
You can use the graph
to find Δx from t = 0 to t = 3 s. Use the data table to check your answer
by determining the displacement from x - x_{0}. What is the
displacement from t = 0 to t = 6 s? If you answer anything other than 0 m,
you should revisit the definition of displacement.
See
Illustration 3.2 for more details on what happens to the acceleration when
the angle of the hill is varied.
Illustration authored by Aaron Titus.
Script authored by Aaron Titus and Mario Belloni.
© 2004 by Prentice-Hall, Inc. A Pearson Company