What parameter of the motion is given by the slope of the velocity vs time curve? How does this value compare to the acceleration calculated in part 2.A.a. Explain.?What parameter of the motion is given by the y-intercept of the velocity vs time curve? Explain.? Part 2: Non-uniform (but constant acceleration) MotionA. Uniform acceleration away from the Sensor:a) For the second part of the laboratory you will essentially repeat the exercises of Part 1, butthis time the data will represent an object undergoing uniform (same everywhere) andconstant (same over time) acceleration. To create a uniformly accelerating cart on the trackraise the end of the track near the sensor slightly using books from the laboratory bookshelf.Start the LQ measurement program, release the cart and let it slide away from the sensor.This time the resulting plot of position vs time is not linear but quadratic (i.e. x varies as t²).Highlight the part of the data representing the moving cart, do a curve-fitting for thatsegment of data using a quadratic model equation (y (x) = Ax² + Bx + C), and present yournumerical results below.5.5442-0.45406Using your data for the fitting parameters and the kinematic equation for position vs time,determine your experimental values for initial velocity and acceleration. Show your calculationsand results here:2 (a)A=11.972ccelerationB =23.944Initial velocity=5.5442 b) Get a printout of position vs time along with the curve-fitting result from one of the localprinters in the laboratory. Be sure your plot fills most of the screen, i.e. the position vs timecurve should include as much of a 3 second recording as possible. Include that printout withyour laboratory write-up. Using that printout pick 5 specific time values within yourmeasurement range, draw a reasonable tangent line to the curve at those 5 points, thendetermine the approximate numerical value of the slope of that tangent line and record thatdata here:Time (s)1.251.301.351.401.45Slope (m/s)1.0151.0461.0TT1.0971.1347From your reading of the chapter on kinematics you should have picked up on the results that theslope of the position vs time curve is the numerical value of the velocity at that time. Make aplot of slope (i.e. velocity) vs time from the data table. You can do it using graph paper or a GCor Excel. The data should suggest a linear relationship between velocity and time, so fromwhatever method you used list the fitting parameter for a linear regression modeling of yourdata.777-.7261.30 -1.25●Slope=1.02Intercept =190What parameter of the motion is given by the slope of the velocity vs time curve? Howdoes this value compare to the acceleration calculated in part 2.A.a. Explain.****What parameter of the motion is given by the y-intercept of the velocity vs time curve?Explain.

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Slope of velocity vs time graph represents the value of acceleration. From graph , slope = 1.02…

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What parameter of the motion is given by the slope of the velocity vs time curve? How does this value compare to the acceleration calculated in part 2.A.a. Explain.? What parameter of the motion is given by the y-intercept of the velocity vs time curve? Explain.?

What parameter of the motion is given by the slope of the velocity vs time curve? How does this value compare to the acceleration calculated in part 2.A.a. Explain.? What parameter of the motion is given by the y-intercept of the velocity vs time curve? Explain.?

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter2: Representing Motion

Section: Chapter Questions

Problem 67A

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