How Does Ticker Tape Reveal Uniform Motion Characteristics?

  • Thread starter Thread starter Karma
  • Start date Start date
  • Tags Tags
    Motion Uniform
AI Thread Summary
The discussion focuses on how ticker tape can illustrate uniform motion characteristics in a lab setting. When walking with non-constant velocity, the dots on the ticker tape will be unevenly spaced, reflecting changes in speed during each time interval. A non-linear position-time graph indicates that the object is either accelerating or decelerating, rather than moving at a constant velocity. If the experiment is conducted at a faster speed, the dots will be spaced further apart, resulting in a steeper slope on the graph, while a slower speed will yield dots that are closer together, leading to a shallower slope. Overall, the spacing of the dots and the shape of the graph provide insights into the motion's characteristics.
Karma
Messages
76
Reaction score
0
ok this is a basic and simple question for UNIFORM MOTION. We did a lab in which a a Ticker tape and ticker tape was used. We were to hold on to the end of a tape while we walk severel steps as our partner operates the ticker timer. As I am walking I am pulling the tape as smoothly and steady as possible.
And then we had to draw a position time graph by measuring and recording the position for each time interval (6 dots, with each space representing an elapsed time of 1/60s, so each time interval consists of 0.10s ((6x 1/60s=0.10s))
I just need help with answering these questions

suppose you were to walk with a non-constant velocity. How might this appear on your ticker tape? Describe and explain the appearance of the dots on the tape.

If you postion time graph were not a straight line, what conclusions could you draw about your motion.

How would the graph change if you repeated the experiment but moved more quickly? more slowly?
 
Physics news on Phys.org
Karma said:
suppose you were to walk with a non-constant velocity. How might this appear on your ticker tape? Describe and explain the appearance of the dots on the tape.

If you postion time graph were not a straight line, what conclusions could you draw about your motion.

How would the graph change if you repeated the experiment but moved more quickly? more slowly?

If you are walking at a non-constant velocity, this means that your velocity is changing, and therefore the rate that you pull the tape through the timer is changing. How do you think pulling the tape through at changing rates will affect the holes?

If your position vs time isn't a straight line when plotted, it must be curved. If its curved, then the gradient is changing. Knowing what you do about what the gradient of position vs time plot represents, what can you say about your motion.
 


If you were to walk with a non-constant velocity, the dots on the ticker tape would be spaced unevenly. This is because the ticker tape records the position of the object at specific time intervals, and if the object is moving with a non-constant velocity, the distance it covers in each time interval will be different. This would result in dots that are closer together or further apart, depending on the speed of the object at that particular time interval.

If the position-time graph is not a straight line, it means that the object is not moving with a constant velocity. This could indicate that the object is accelerating or decelerating, or that it is moving in a curved path.

If the experiment is repeated with a faster speed, the dots on the ticker tape would be spaced further apart as the object is covering more distance in each time interval. This would result in a steeper slope on the position-time graph, indicating a higher velocity. On the other hand, if the experiment is repeated with a slower speed, the dots on the ticker tape would be closer together, resulting in a shallower slope on the position-time graph, indicating a lower velocity.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top