Thread: Quadratics

Can you please help me with this question

Point O is the intersection of 2 roads which cross at right angles; one road runs from north to south, the other from east to west. Car A is 100 m due west of O and travelling east at a speed of 20m/s, and Car B is 80 m due north of O and travelling south at 20 m/s.

a) show that after t seconds their distance apart, d metres is given by
d^2 = (100-20t)^2 + (80-20t)^2

b) show that this simpifies to

d^2 = 400((5-t)^2 + (4-t)^2 ).

c)show that the min distance apart of the 2 cars is 10 squareroot 2 m

At any point in time, car 1 is positioned at and car 2 is positioned at .

If you were to draw a segment between these two points and create a right-angle-triangle, you can evaluate the distance between the two points using Pythagoras.

How can i get (5-t)^2 + (4-t)^2
from 2t^2 -18t +41

You don't. What you should be doing is taking out common factors BEFORE trying to expand.

what I got was
d^2 = 800t^2 - 7200t +16400
I took 400 as the common factor and got
d^2 = 400(2t^2 -18t +41)

Was what I did wrong ??

Mathematically it all appears correct. But your APPROACH is wrong.

Surely you can see that . When you are trying to get to a factorised form, the best approach is to factorise...

Originally Posted by Prove It

Mathematically it all appears correct. But your APPROACH is wrong.

Surely you can see that . When you are trying to get to a factorised form, the best approach is to factorise...

oh!
okay
thnks

in the last part this is what i did:

d^2 =400(2t^2 +41 -81t)
d^2 = 800(t-9/2)^2 + 200
d^2 = 200
d= 10 squareroot 2

Is my approach correct or is there a faster way of solving it than completing the square??

Yes your approach is correct. Well done.

Originally Posted by prove it

yes your approach is correct. Well done.

thanks !! ^__^