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Differentiation of displacement and time to get velocity

  • Thread starter Ocis
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  • #1
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[SOLVED] Differentiation of displacement and time to get velocity

A moving body is related to a fixed point by the displacement equation: s = 12( 1 - e^-t).
Assuming the body is moving in a straight line on a flat plain.

1. obtain an expression for the velocity after time t seconds.

2. obtain an expression for the acceleration also after t seconds

3. Calculate the time at which the velocity of the body is 6 ms-1.




So i know that s = v t. So v = s/ t and ds/dt = v



My attempt has gone round in circles because I am unsure of how exactly to differentiate it do I ignore the 12 first and diff. it, or times out the bracket then diff. it....? Too confused..
 

Answers and Replies

  • #2
nicksauce
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Well to simplifiy s = 12 - 12e^(-t)

What if the derivative of a constant? What is the derivative of an exponential?
 
  • #3
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constant = 0 and derivative of an exponential would be - e ^-t so its just 12 e ^-t ?
 
  • #4
nicksauce
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I would agree.
 
  • #5
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but then to differentiate it a second time it becomes what - 12 e^-t ? how do I work out what the time is then for part 3. i would have e^-t = 6/12 ? I am stuck where to go from there how do I do it?
 
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  • #6
nicksauce
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Do you know how to use logarithms?
 
  • #7
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I have in the past but am not overly keen on them. Would it be something along the lines of
-t log e = 6/12 and then 6 /12 log e = -t well that looks wrong, what else..?
 
  • #8
nicksauce
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e^-t = 6/12
-t log e = log(1/2) (taking the log on BOTH sides)
and since we know log e = 1 (when using base e, ie the 'ln' button on your calculator) and knowing that log(1/2) = -log(2), we get
t = ln 2

It might be a good time to remind yourself about all the properties of logarithms, as they are mucho important.
 
  • #9
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Yeah you are right, I do - many thanks. Is the second derivative right that i wrote earlier?
 
  • #10
nicksauce
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Yes it is right.
 

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