Register to reply 
Ticker Timer help 
Share this thread: 
#1
Apr1506, 09:41 PM

P: 161

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 5tick 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 ^^" . 


#2
Apr1506, 11:21 PM

P: 16

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! 


#3
Apr1606, 02:12 AM

P: 161

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 ?



#4
Apr1606, 04:06 AM

Emeritus
Sci Advisor
PF Gold
P: 9,772

Ticker Timer help
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; [tex]a = \frac{dv}{dt} = \frac{\Delta y}{\Delta x}[/tex] 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 


#5
Apr1606, 07:24 AM

P: 161

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 ?



#6
Mar3111, 10:22 AM

P: 1

I'm new here! Apologize for my bad english ><
I'm doing similiar 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 (71) = 1.2s. My question is why minus one? Where did the time taken for one strip gone?? Your answer will be much appreciated. 


Register to reply 
Related Discussions  
Uniform Accelerated Motion  Ticker Tape Experiment  Introductory Physics Homework  2  
Ticker tape experiment  Introductory Physics Homework  2  
About tickertape timer  Introductory Physics Homework  2  
555 Timer  Electrical Engineering  3  
555 Timer Help  Electrical Engineering  5 