Question

The following data table gives the position of an object during freefall. The time interval between adjacent points is 1/60 seconds.

Point	Position(cm)
1	94.30
2	91.22
3	87.87
4	84.25
5	80.35
6	76.19
7	71.75
8	67.03
9	62.05
10	56.79

What is the instantaneous velocity (in cm/s) of the object when it is at position 9?

Answer 1

Solution:

We can plot the graph in the MS Excel or in any other graphing tool. The following is the graph of the data point given in the problem. The points in the graph are connected with a polynomial trendline and the polynomial of the degree 2 is shown.

The equation y = -490.64x² - 160.1x + 97.103 gives us the position of the object during free fall for every value of the x. The x values are the time and y values are the position of the object in cm

The derivative of the above equation with respect to x (i.e time) gives us the equation for velocity.

v = dy/dx = d(-490.64x² - 160.1x + 97.103)/dx

v = dy/dx = -981.28x – 160.1

Thus instantaneous velocity at point 9 that is at time x = 0.15 s is given by plugging x = 0.15 s in above equation of velocity

v _{at point 9} = v|_0.15s = -981.28*0.15 – 160.1

v _{at point 9} = v|_0.15s = -307.29 cm/s

Thus instantaneous velocity of the object at point 9 is -307.29 cm/s