If an object has a displacement function #s(t)=t-ln(2t+1)# where #t# is in seconds and #t>=0#, can you find the distance travelled in the first 2 seconds?

1 Answer
Eddie
Mar 19, 2017

#= 3/2 + ln (2/5) #

Explanation:

Displacement is given by:

#s(t)=t-ln(2t+1)#

We can see that the change in displacement (the vector quantity that measures spatial differences) during the period #t in [0,2]# would simply be:

#s(2) - s(0) = 2 ln 5#.

But that is not necessarily the distance (the scalar quantity that measures spatial differences) travelled by the object in #t in [0,2]#.

Solving the displacement equation looks tricky, ie we need to solve:

#t-ln(2t+1) = 0#

I am not sure how you do that simply, so, despite the Pre-Calc classification, from calculus , we can see that velocity is:

#v(t) = dot s(t) =1-2/(2t+1)#

#v(t) = 0 implies 1-2/(2t+1) = 0 implies t = 1/2#

So, just to make it clear:

#v(0) = -1#, object is moving to left

#v(1/2) = 0#, object has stopped moving

#v(2) = 3/5#, object is moving to right

To obtain the distance travelled, we need to look at the displacements at these 3 points:

#s(0) = 0#, object is moving to left

#s(1/2) = 1/2 - ln 2 approx -0.2#

#s(2) = 2 - ln 5 approx 0.4#, object is moving to right

So the distance #s_(0,2)# moved during the period #t in [0,2]# is:

#s_(0,2) = 2 - ln 5 + abs(1/2 - ln 2)#

#= 3/2 + ln (2/5) #

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