1-Using your graph in the Data section above, a ruler and the trendline from Par

Question

1-Using your graph in the Data section above, a ruler and the trendline from Part 1 (but not the equation of the trendline), extrapolate the position of the car when t =6.27 s. Draw the lines you used to obtain your answer on your graph above but give your answer with correct units here. 2- Using the equation of the trendline from the graph in Part 1, determine the position of the car when t=6.27 s. Show all of your work below (remember to include units in your calculations). Record your answer with correct units and significant figures. 3-Determine the position of the car at t = 0 s using two different methods and show all work needed below. Be sure to record your answers with correct units and significant figures. Discuss which method is more accurate and one reason why it is more accurate. Be specific

Part 1 Making Tables and Graphs: Graph Time(s Distance(cm) Distance(m) Velocity(m/s) V(miles/hr 5.42 2.51 1.82 2.59 3.97 5.4 6.76 7.38 8.24 9.49 633 912 1313 1648 1987 2074 2327 2595 6.33 9.12 13.13 16.48 19.87 20.74 23.27 25.95 Part 1 Making Tables and Graphs: Data Table Distance(m) 30 y-2.5097x + 2.5404 25 20 15 10 0 0 4 6 10

Explanation / Answer

Draw a vertical line at X= 6.27 that is the time and find the point which is cut by this line with the curve. From this point draw a horizontl line to Y axis. This wil give you the distance car has moved within the given time.

So answer is approximately 18 m.

So you have the equation of the straight line from distance VS time(Normal graph will be named as Y vs X) as

Y = 2.5097x+2.5404

you want to find distance ie, Y at time ie, X = 6.27.So put the value of X in the equation and find out Y.

Y = 2.5097* 6.27+ 2.5404 =18.276m

3) You can use the same graph and equation to find out the position of the car at T=0. Basically you need interpolate the straight to y axis using a ruler and see where it cut the Y axis which is approximately 2.5 M. In other words when you have an equaiton of straight in general Y = m * X + C

m is the slope of the straight line in this case it is nothing but he velocity

C is the Y intercept. So compaering the general eqn with the given C = 2.5404m

So at t=0 Car is at 2.5404m

Another way to find it out again in the given equation put X ie, t=0.This will give you Y = 2.5404m

The answer will be more accurate and correct when you extract the data with equaiton when you do manally with ruler it can give you error and you can't give an answer correct to 3 significant figures.

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