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avatar+15 

This one has been tripping me up, any help?

 Sep 17, 2019
 #1
avatar+104869 
+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

 

cool cool cool

 Sep 17, 2019
 #2
avatar+15 
+1

Oh right thanx!

Soapen  Sep 17, 2019
 #3
avatar+19822 
+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.....

 Sep 17, 2019

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