a.    From my real life I will be using the speed of train starting from a station.

b.    In this experiment a train started from a station where it was stationary and starts moving as time increase its speed increases with time. The recorded data is as follows.

c.    Please find the table below

TIME(Sec)	0	2	5	10	14	27	40
SPEED(m/s)	0	4	13	20	27	40	48

d. The graph of the table is below

e.    In the above graph the three secant line is can be obtain from these four points

Let points A(5, 13), B(10, 20)  C(14, 27) and D(40, 48)

For the line AB the slope will be

M1 = (20 – 13) / (10 – 5)

      = 1.4

           For line AC the slope will be

M2 = (27 – 13) / (14 – 5)

      = 1.55

For line AD the slope will be

M3 = (48 – 13) / (40 – 5)

      = 1

f. As a result the graph with the secant line will be shown as

g.    From the graph it is easy to see that the speed of the train goes up with different rate. To find instantaneous rate of change of speed with respect to time can be calculated by the secant line drawn in the above graph by taking another point Q on the secant line. Find the slope of line joining point A and Q. the point A is one through which all the secant line has passed. This slope will give the rate of change of speed with respect to the time.

We can see there are different rate of change of speed for the intervals. The calculation of rate of change of speed with respect to time will be the acceleration of the train.

h. The IRC at a point on the given graph will give the instantaneous acceleration of the particle at given point. However, the slope calculated above gives the average acceleration for the given interval.

In the calculation of IRC or derivative at a point we have taken ratio of change in speed to the time interval. Hence the unit of IRC or derivative at a point in this case will be meter/second². From the calculation we have seen that the rate of change of speed with respect to time between t = 5 to t= 13 sec.  Is 1.55 m/s. This shows the average acceleration between these to time instant.

Hence, acceleration of the train at time t = 22 sec will be 1.55 m/s².