Tangent Line Problem

The instantaneous velocity at any point will simply  be the slope of the line through that point

For instance....look at the line through the point that represents the position after 2 sec

The following approximate points are on this line  (1,0)  and ≈ ( 3.5, 60)

So....the approximate slope  is  [ 60 - 0 ] / [ [3.5 - 1 ] =  60 /2.5   ≈  24m / s

And look at the line through the point that represents the position after 4  sec

The following approximate points are on this line  (2,0)  and ≈ (4.5, 120)

The approximate slope is   [120 - 0 ] / [ 4.5 - 0  ] = [120/4.5]  ≈  26.67 m/s

This makes sense.....the slope at 4 sec  is steeper than at 2 sec....so....we would expect the instantaneous velocity at 4 sec should be  > than at 2 sec

I think tangent lines are drawn on your graph at these points......the SLOPES of the lines will be the velocity at that point.

At 2 seconds you have a line with slope   rise/run     50/2 = 25 m/s    approx

at 4 seconds the tangent line is 95/2 = 47 m/s    approx.

    going faster after 4 seconds due to acceleration.....