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# Illustration 2.5: Motion on a Hill or Ramp

Animation 1 | Animation 2 | Animation 3

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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.

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