Graph and Questions: Solve Motion of 500.0 g Cart

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In summary, the conversation discusses the equations and units used to calculate motion, specifically the average speed and distance traveled by a cart. It also explains how to convert the mass of the cart from grams to kilograms and how it affects the cart's motion. The mass affects the force and acceleration of the cart, but not its average speed.
  • #1
BrownBoi7
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The graph in the picture attached describes the motion of a 500.0 g cart for the first 60 seconds of its motion.
1. Calculate how far the cart moves during the fi rst 25 seconds of motion.
2. Indicate the time intervals during which the cart has a non-zero acceleration.
3. Calculate the acceleration of the cart at 35 seconds.
4. Indicate all times during which the cart is at rest.
5. Calculate how far the cart is from its starting location after 60 seconds.
6. Calculate the additional time required for the cart to arrive back at its initial position if it continues to move at 50 m/s as indicated on the graph.

My attempt:
1. area of triangle + area of rectangle = total displacement in first 25 seconds = distance
1000 m + 500 m = 1500 m
2. (0,20) and (30,45)
3. 50-100/5 = -50/5 = -5 m/s^2
4. At t = 0 and t = 40
5. Total area covered? 1000 m + 1000 m + 500 m + 125 m + 750 m = 3375 m
6. I did not understand this question. Do i just get the area (distance) and speed is given. So i can get the time taken? I don't know why they would give the mass of the object?
I haven't used it answering any of the questions--that kinda puts me in doubt as to if I answered them correctly.

Can somebody please review them and let me know where I'm wrong?

Thanks!
 

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  • #2
You have the right idea for all of these I think. 1 - 4 are definitely right.

For 5, it will be the total area covered, but you have to make the areas below the t-axis negative, since those are places where the cart moves backwards, which would move it closer to the starting location. If it was asking for the total distance traveled you would make each area positive, but for displacement you include signs on all the areas. Those questions can be difficult to answer.

For 6, I think you need the answer from 5. For 5 you get a positive area representing how far the cart is from the starting point, so 6 is essentially asking how much "negative" area you would need to cancel out the area in 5 completely, if you have the cart moving at -50 m/s as in the graph. You could also use the displacement numerically, and then do a simple calculation with constant to velocity to find how much time is needed to cover the displacement, in the opposite direction of course.

I don't see a use for mass in any part of the problem, so don't worry about it. Sometimes they'll throw that in there just to trick you, or for the sake of completeness in the problem description.
 
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  • #3
@ jackarms You always answer my questions :) Thank you bro.
Anyways, getting back to the question. I calculated the total distance traveled by the cart. So displacement is direction sensitive, so to speak. Would it help if I converted it into a position vs time graph? I don't think it is needed but I'm still a wee bit confused on the negative displacement part (below the x axis) It does kinda make sense so here's my attempt at 5:
2500 m - 125 m -750 m = 1625 m
In other words, the cart is 1625m from it's starting position?
And now we can use this to find 6:
D= SxT ; T= D/S
1625 m/ -50 m/s = 32.5 secs

Does that sound right?
 
  • #4
Also, I think my original answer to 2. is wrong. Well there's nothing wrong with (0,20), however, (30,45) should be broken down into two parts because the velocity is zero at t=40, which means the cart was at rest for a second?Should I break it into (30,39) and (41,45)? or simply (30,40) and (40,45)?

This is so confusing.
 
  • #5
And did I mess up + - for 3?
Could it be -50m-100m/5 secs = -150 m/5 secs= -30 m/s^2?
My reason is since the curve is downwards the velocity must be decreasing--so it's -50 m/s at t=35 instead of 50 m/s
 
  • #6
Hey, no problem for answering your questions. It's always nice to brush up on my physics too.

Your calculations of displacement and the time for #6 both look right. And you're right about displacement being direction sensitive. That's the difference between it and distance traveled, and they're certainly easy to mix up. Making a position-time graph is another possibility, but it's difficult since the curve would be parabolic, and to determine all the points you would have to do the area calculations anyway. If you can understand the concepts of positive and negative area for a velocity-time graph, I think that would be easier.

For #2, your original answer is correct. It's a little confusing how the cart can still be accelerating when it's at rest, but it still is, so you would want to include t = 40 in the answer. You can think of it as the nonzero acceleration changing the velocity from positive to negative just at that point. Another example is if you toss an object up into the air, what will be the acceleration at it's maximum height. It's tempting to say zero since it's at rest, but it's still in free-fall, so it must still be accelerating.

And for 3, your original answer is also right. It's just the slope of that line, which you correctly found. Even though the velocity is decreasing at t = 35, the velocity is still positive. If you wanted to use the -50 m/s value, you would have to use the time interval 15 second time interval from 100 m/s to -50 m/s. I misread that the first time I looked at it too.
 

Related to Graph and Questions: Solve Motion of 500.0 g Cart

1. What is the equation for calculating motion?

The equation for calculating motion is motion = distance / time. This equation is also known as the average speed formula and is used to determine the average speed or velocity of an object.

2. How do you calculate the distance traveled by the cart?

To calculate the distance traveled by the cart, we need to know its average speed and the time it took to travel. We can then use the equation distance = average speed * time to find the distance traveled.

3. What is the unit of measurement for distance and time in this problem?

The unit of measurement for distance in this problem is meters and for time is seconds. These are the standard SI units used for measuring distance and time in scientific calculations.

4. How do you convert the mass of the cart from grams to kilograms?

To convert the mass of the cart from grams to kilograms, we need to use the conversion factor 1kg = 1000g. This means that for every 1000 grams, there is 1 kilogram. So, to convert from grams to kilograms, we divide the mass in grams by 1000.

5. How does the mass of the cart affect its motion?

The mass of the cart affects its motion by determining the amount of force needed to move the cart and the acceleration it can achieve. A larger mass will require a greater force to move and will have a slower acceleration compared to a smaller mass. However, the mass does not affect the average speed of the cart.

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