how do you find the maximum distance of a trajectory? (1 Viewer)

[poptarts12345]

How would I do this question?
A particle moves according to x=t^2-8t+7. What is the maximum distance from the origin and when does it occur :
a) during the first 6 seconds
ANS : 9m when t=4
b) during the first 10 seconds
ANS : 27m when t=10

 

[B1andB2]

I think that with these max/min qs you have to find the vertex which is -4 sub it then back in? Pls correct me if i am wrong

 

[TheShy]

easiest thing to do is to draw a rough sketch of the graph. Then you can easily find the max distance with a given time. For a) you can imagine that you're restricting the domain so that its only from -infity to 6 (As it asks for the first 6 seconds). then you just need to find when it is the highest. Same thing for part B

 

[Trebla]

Since it is about distance (from the origin) and not displacement, a good approach is to try to sketch the path of particle on the number line over time. Initially, it’s at x=7 and then it moves in the negative direction to x=-9 before it turns around and moves in the positive direction.

 

[CM_Tutor]

Remember that the object can only change direction when it stops, ie when dx/dt = 0.

Maximum distance from O can thus only be when the particle stops or at the start or end of the motion.

Here, dx/dt = 0 gives t = 4, at which time x = -9

At the start, t = 0, and x = +7

At the end, t = 6, and x = -5.

The furthest from O is thus x = -9, a distance of 9 units, at t = 4.

Extending to t = 10, we get x = 27.

So, on [0, 6], maximum distance = |x|_max = 9

and, on [0, 10], maximum distance = |x|_max = 27

On a number line:

<--(-10)---------(-5)---------(0)---------(5)---------(10)---------(15)---------(20)---------(25)---------(30)--> x
Stop / turn at -9 (at t = 4) . . . . . . . . . Start (t = 0) at +7
. . . . . . . . . |-----------<------------<------------* START
. . . . . . . . . |--->----*-------->--------------->------------------->---------------->----------------* END
. . . . . . . . . . . . . . . . * pass x = -5 at t = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . continue on to reach x = +27 at t = 10