#1**+2 **

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

CPhill Sep 17, 2019

#3**+1 **

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.....

ElectricPavlov Sep 17, 2019