velocity (1 Viewer)

[stressedadfff]

hey guys can someone please explain the following part d

 

[specificagent1]

distance is equal to the total area under the velocity curve, you may not be able to run it as one integral from 0 to 3 as there is postive and negative areas

 

[stressedadfff]

distance is equal to the total area under the velocity curve, you may not be able to run it as one integral from 0 to 3 as there is postive and negative areas

oh how would you suggest then

 

[specificagent1]

oh how would you suggest then

so the first derivative (velocity) is 2 - 3/(t+1) right?

and the graph looks like this:


Essentially, we want the area under the curve from 0 to 3 so we would run the first | integral from 0 to 0.5 | (absoluted because of negative areas) + integral from 0.5 to 3

 

[CM_Tutor]

The faster way is to consider the position of the particle at the start and finish of the period of interest, plus at any times in the interval when it stops (hence part (c)) as it may change directions.

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From part (c), we know that the particle is stationary at , so check

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So, the particle's travel path:

• 
• 
• 

And so, over this period of 3 s,

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while its total displacement is .

Using the integration approach, you should find:

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