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In the tutorial about velocity and acceleration graphs last week, most of you didn’t finish the last set of questions about a cart going up and down a ramp. The first two pages below cover that same tutorial material. If you didn’t complete all of the tutorial questions do them below, working with a group, if possible. DO NOT hand in these first two (tutorial) pages with this homework .

Up and down a ramp

In all these questions, just think about the cart’s motion after you stop pushing it up the ramp but before you catch it on the way down.

1 Velocity graph review.

1 This graph shows the cart’s velocity vs. time. On those same axes, graph the cart’s speed vs. time. Speed is how fast something is moving, independent of direction.

[Speed shown in gray]

2 What would you guess is a common mistake students make on the velocity vs. time graph for this and similar situations? Explain.

Graphing speed instead of velocity, i.e., not taking into account the direction of motion.

2 Acceleration on the way down

1 Consider the cart as it rolls down the ramp after reaching its peak. Give a reason why a smart student might think the cart’s acceleration during that segment of the motion is positive. Briefly summarize the reasoning.

Since it’s speeding up, the acceleration is positive. If it were slowing down, the acceleration would be negative.

2 Now give a reason why a smart student might think the cart’s acceleration during that segment of the motion is negative. Briefly summarize the reasoning.

The velocity is still going “down,” as the velocity graph shows. Also, motion towards the detector “flips” the sign of the velocity, so it makes sense that it would also “flip” the sign of the acceleration.

3 Below, sketch two different predictions for the cart’s acceleration vs. time. On "Prediction graph 1," assume the cart’s acceleration as it rolls down the ramp is positive (argument 1 above). On "Prediction graph 2," assume the acceleration is negative during that segment (argument 2).

3 When performed with good equipment, the acceleration graph comes out as shown here. Two students disagree about what they should do next to continue learning about acceleration.

VENUS: Look, now we know which argument to believe about whether the acceleration is positive or negative when the cart rolls down the ramp. Physics is an experimental science! Since we know the right answer and the reasoning behind it, we’re ready to move on.

SERENA: But we also came up with a sensible argument that the acceleration would be positive, not negative. To understand acceleration, I think we also need to understand what’s wrong with that argument.

Which student do you agree with more in this case? Explain.

Good arguments can be made either way. Here’s one possible compromise. To take Serena’s idea into account, Venus could say that, in this particular case, the explanation of why acceleration is negative coming down the ramp already contains, “built into it,” a fairly clear explanation of why the acceleration isn’t positive.

4 How to decide?

1 Consider the contradictory arguments from part B about whether the cart’s acceleration is positive or negative as it rolls down the ramp. Was it possible to decide for sure, without doing the experiment, which argument was correct? Explain.

You can decide the issue by checking for coherence between the velocity and acceleration graphs. The velocity graph is negatively sloped the whole time, including when the cart rolls back down the ramp. So, the acceleration is negative the whole time.

2 Let’s see if the president-for-life of mistake-avoidance strategies can help us here. Look at the velocity graph on the previous page.. By checking for coherence between your velocity graph predictions and your acceleration graph, could you decide whether the acceleration is positive or negative as the cart rolls down the ramp? Explain. (If your above answers addressed this issue, skip this question.)

See above answer.

3 We just saw that checking for coherence can help you decide between sensible competing arguments. Can that game also help you understand why the other argument is wrong? See if you can use the connection between velocity and acceleration graphs to explain, in a common sense way, why the cart’s acceleration is negative even though it’s gaining speed (as it rolls down the ramp)? Hint: Is the cart’s velocity (not speed) trending “up” or “down?”

Intuitively speaking, acceleration is positive when the velocity goes up and negative when the velocity goes down. As the velocity vs. time graph for this experiment shows, the velocity is still “going down” when the cart rolls down the ramp. In general, when something negative gets more and more negative, it’s “going down.” For instance, if your bank balance is negative (meaning you owe the bank money!) and it keeps getting more negative, your net worth is going down, not going up!

Rolling ball

A ball, released from rest, rolls down ramp A, then along the floor, then up ramp B, as drawn here.

1 Sketch the ball’s velocity vs. time and acceleration vs. time, from the moment it’s released until the moment it reaches its peak on the second ramp. Label your graphs in such a way that we can tell when the ball is on ramp A, when it’s on the floor, and when it’s on ramp B. Remember to end your graphs at the moment the ball reaches its highest point on ramp B.

3 Now sketch distance vs. time. (“Distance” here means the distance the ball has rolled along the track from its starting place.) Explain your reasoning in words.

On ramp A, the ball gains speed, which means the slope of the position graph increases; the ball covers distance at an increasing rate. On the floor, the ball “keeps” the high speed it attained and rolls at constant velocity, indicated on the position graph by a constant slope (straight line). On ramp B, the ball slows down but doesn’t reverse direction (until after the graph ends). So, the position graph keeps going up, though at a decreasing rate (slope).

4 Bill got his position and acceleration graphs right but drew the incorrect velocity vs. time graph shown here.

1 Why do you think Bill make this mistake? What advice would you give him to help him avoid that mistake in the future?

He seemed to be thinking that something about the motion “stops” on the floor. That’s true about the acceleration, not the velocity. When he wants to indicate that something “stops,” Bill should think carefully whether the object stops moving (zero velocity) or stops speeding up or slowing down (zero acceleration).

2 After drawing his velocity graph, how could Bill have realized that it’s wrong? (If you already answered this in question 1, just say so.)

Checking for coherence between the velocity and acceleration graphs would not catch the mistake here. Even though Bill’s velocity graph is wrong, it “fits together” with the acceleration graph because the slope of his velocity graph in each segment does indeed equal the ball’s acceleration. (Make sure you understand why.) However, Checking for Coherence between velocity and position would catch the mistake, because Bill’s velocity graph doesn’t fit together with the position graph.

Velocity graph

Consider the velocity vs. time graph shown here, focusing on the issue of whether the object turns around (reverses direction).

1 Why might a smart student say the object reverses direction?

The graph switches from trending up to trending down—a switch in direction.

2 Why might a smart student say the object doesn’t reverse direction?

The velocity stays positive the whole time, meaning the object always moves away from the detector.

3 Does the object reverse direction? As part of your explanation, explain why the other argument is incorrect, and give advice to someone who made that mistake that can help them understand why they made the mistake and how they can avoid it in the future.

Argument B is correct. The object “switches” from speeding up to slowing down, but it keeps moving in the same direction. If this were a position graph, it would indicate a reversal of direction. When someone sees a slope “reversal” on a graph, he or she needs to think carefully about whether the reversal is from speeding up to slowing down (as is the case on a velocity graph) or a reversal in the direction of motion (as is the case on a position graph).

4 Sketch a graph of the object’s position vs. time.

Bouncing ball

A hard rubber ball is dropped from rest. It falls to the concrete floor and bounces back up almost to its initial height. A motion detector is mounted on the ceiling directly above the ball, facing down. So, in this example, the positive direction—the away-from-the-detector direction—is downward.

1 On these graphs, sketch the ball’s position, velocity and acceleration vs. time. Treat the bounce as a sudden event that happens in a single instant of time. (Below, you’ll think about what’s going on during the bounce.) Because you’re dealing with all three kinds of graphs at once, think carefully and use the strategies you’ve learned for avoiding and catching mistakes.

2 Two students discuss the acceleration during the bounce, i.e., during the brief time the ball is in contact with the floor.

SMITHA: The ball’s acceleration must be large during the bounce, larger than when it’s just falling or rising. Because during the bounce, the ball’s velocity changes a lot, very suddenly, a big change in velocity over a very small time. So, the acceleration — the rate of change of velocity — is big.

AJITA: But look...because the ball is very bouncy, it hardly loses any velocity during the bounce. That’s why it rebounds almost to its initial height. For instance, if it reached the floor going down at 8.0 meters per second, then maybe it bounces back up at 7.9 meters per second or even 7.95 meters per second. So, during the bounce, the velocity changes by a tiny amount. Even though this change happens over a small time, the change in velocity per change in time isn’t too big, probably smaller than when it’s just rising or falling.

Which student do you agree with? Explain.

Smitha is right. Ajita is talking about the change in speed, not the change in velocity. At the bounce, the velocity suddenly jumps from a positive value, such as +8.0 m/s, to a negative value, such as –7.9 m/s. The change in velocity, is therefore ∆v = vafter bounce – vbefore bounce = –7.9 m/s – 8.0 m/s = –15.9 m/s. That’s a large change in velocity to happen in such a short time. So, the acceleration is large.

3 During the bounce, is the acceleration positive or negative? Explain.

Negative. As the above velocity vs. time graph shows, during the bounce, the velocity “jumps” down, switching from positive to negative. So, that part of the velocity graph has a (huge) negative slope.

The same conclusion pops out of the calculation we did in part C. Since ∆v is negative, so is the average acceleration, = ∆v/∆t. Make sure you can hook up the graph and math.

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