Ticker Timer Experiment: Calculating Displacement, Velocity & Acceleration

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The discussion revolves around conducting a ticker timer experiment to calculate displacement, average velocity, and acceleration of a trolley. Participants clarify that total displacement can be found by subtracting the initial position from the final position, while average velocity is determined by the slope of the position graph. For acceleration, it's noted that it can be calculated as the change in velocity over time, and the area under a velocity-time graph represents total displacement. A participant expresses confusion about their graph lacking values, seeking guidance on how to derive numerical data from the tape strips. Additionally, there is a query regarding the time calculation for multiple strips, specifically why the total time is not simply the sum of individual strip times.
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Hi , I'm new here =) .
This is from my homework . My teacher let us do the ticker timer's experiment with 50 GHz . after that , we r suppose to calculate the total displacement and the average velocity and also the acceleration of the trolley . we cut the tape into 5-tick strips and pasted into a graph . The graph show ... err a constant acceleration i guess , coz it's going up . But then how am i suppose to calculate the total displacement and average velocity ?

Please help =) .
sorry for my bad english x( , I'm from malaysia ^^" .
 
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Are you sure the acceleration is constant? There may be acceleration, but if the velocity graph does not appear linear then it will not be technically "constant acceleration."

To find the total displacement, I think you would just need to subtract the initial position value from the final position value.

Average velocity would be equal to displacement over time elapsed. (On your graph this would be equal to the slope of the position graph).

Also, you mentioned you will need to calculate the acceleration of the trolley. I assume you mean the average acceleration? If this is true, then you would want to subtract your initial velocity from your final velocity and divide by total time elapsed.

I hope this helps!
 
thanks , i get it . But , my graph have no value , how do i get the value to substitue ? do i measure the tape's length ?
 
Welcome to PF Crays,

If you are plotting a velocity - time graph (I assume you are) then the acceleration would be given by the gradient of the the line. Remember acceleration in the change in velocity over time hence;

a = \frac{dv}{dt} = \frac{\Delta y}{\Delta x}

To determine the displacement from a velcoity time graph, you need to find the area under the line, this is equal to the total displacement. If you velocity is a straight line (as it should be with constant acceleration) you can form a triangle using the slope of the hypotenues, with the time axis as the base and the velocity as the perpedicular height.

Regards
~Hoot
 
oh , i get the acceleration part . Still , my graph have no value at all x( . no value at y-axis , not at x-axis . no nothing , i only got strips with different length , each strips have 5 dots in it . How do i get some numbers into the graph ?
 
I'm new here! Apologize for my bad english ><
I'm doing similar experiment and my teacher asked us to calculate the time taken,t.
For one tick is 0.02s, each strip contain 10 ticks, that make it 0.2s for one strip. There were 7 strips which 10 ticks on each strip. However, the time taken for all 7 strips is not 0.2s x 7 = 1.4s, instead it is 0.2s x (7-1) = 1.2s.
My question is why minus one? Where did the time taken for one strip gone??
Your answer will be much appreciated.
 
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