Motion and differentiation question (1 Viewer)

[ataratar]

Hey guys, this probably isn't very hard but I'm very confused! Any help would be much appreciated
A particle is moving in a straight line so that its displacement xcm over time t seconds is given by x=t(square root: 49-t²)
How far does the particle move altogether?
Sorry for the confusion about the square root sign, basically just the 49 - t² is all under a square root multiplied by t on the outside!
Thankya

 

[bokat]

The time has to be between 0 and 7, otherwise sqrt(49-t^2) will not be defined and time can not be negative.
At t=0 x=0 and t=7 x=0. And x is positive as both t and sqrt(49-t^2) is positive.
So if we sketch x versus t it will be from zero to a max and then back to zero.
To find that max differentiate x with respect to t which gives sqrt(49-t^2)-t^2/sqrt(49-t^2) and equate it to zero.
You will get t=7/sqrt(2) sub back in x to get x=49/2. This is the max value for x.
So the answer is 0<=x<=49/2.

Reza Bokat
www.cambridgecoaching.com.au