Differentiation of displacement and time to get velocity

[Ocis]

[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..

 

[nicksauce]

Well to simplifiy s = 12 - 12e^(-t)

What if the derivative of a constant? What is the derivative of an exponential?

 

[Ocis]

constant = 0 and derivative of an exponential would be - e ^-t so its just 12 e ^-t ?

 

[nicksauce]

I would agree.

 

[Ocis]

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?

 

[nicksauce]

Do you know how to use logarithms?

 

[Ocis]

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..?

 

[nicksauce]

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.

 

[Ocis]

Yeah you are right, I do - many thanks. Is the second derivative right that i wrote earlier?

 

[nicksauce]

Yes it is right.