T-C = ((40-4t)-(-10+45t),(20+75t)-(40+70t))
= (50-49t,-20+5t)
at t=0, then
d=|T-C| = √(50^2+20^2) = √2900 = 53.9
d^2 = (50-49t)^2 + (-20+5t)^2
= 2426t^2 - 5100t + 2900
Not sure how you got your expression
the position of a car in time t (in hours) is (-10+45t,40+70t) and a truck (40-40t,20+75t)
find the distance between the car and the truck at time t=0 in km)
i have 53.9km to 1dp
find an expression for d^2 in terms of t)
I have 7250t^2 -8700t+290
Complete the square which I got 7250(t+3/5)^2 +290
and hence show the minimum value of d^2 is attained when t=3/5. However when i plug in t=3/5 I get 10730
Where have i gone wrong? :(
2 answers
Should be (t-3/5)^2 then it should be clear from then. Multiplying by 0 if t=3/5.
1 answer