A train leaves the station at 10:00 and travels due south at a speed of 60km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two closest together?
totally stuck on this one.
A train leaves the station at 10:00 and travels due south at a speed of 60km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two closest together?
totally stuck on this one.
let $\displaystyle d_1$ = distance first train is from the origin (station)
$\displaystyle d_1 = 60t$
$\displaystyle d_2$ = distance second train is from the same station
$\displaystyle d_2 = 45 - 45t$
let $\displaystyle r$ = straight line distance between the two trains between 10:00 and 11:00
$\displaystyle r^2 = d_1^2 + d_2^2$
minimize $\displaystyle r^2$ and $\displaystyle r$ will also be minimized.
Ok, can you or someone check my work below because I can't seem to get the answer with the back.
I said..
Let x be the distance between A and B in km
Let a and b be the distance travelled by A and B in km
Min x at t=?
V=d/t
For A
a=60t
For B
b=45t
b^2+a^2=x^2
(45t)^2+(60t)^2=x^2
x=(5625t^2)^0.5
x'=5625t/[(5625t^2)^0.5]
x'=0 when t=0
domain of t
10<=t<=11
x(10)=750 km
x(11)=825 km
therefore the min distance between train A and train B is 750 km at 10:00
BUT>>> the back of the book's answer is: t=0.36 h
How do i get this and what did I do wrong?
take another look at my original post ...
$\displaystyle r^2 = (60t)^2 + (45 - 45t)^2
$
minimizing $\displaystyle r^2$ will minimize $\displaystyle r$ ...
take the derivative of the RHS and set = 0
$\displaystyle 2(60t)(60) + 2(45-45t)(-45) = 0$
$\displaystyle 3600t - 2025 + 2025t = 0$
$\displaystyle 5625t = 2025$
$\displaystyle t = \frac{2025}{5625} = \frac{9}{25} = .36$ hr
the distances, $\displaystyle d_1$ and $\displaystyle d_2$ are measured from the origin.
the first train starts at the origin and moves away from the origin at 60 km/hr
$\displaystyle d_1 = 60t$
the second train started 45 km away from the origin, and heads back toward the origin at 45 km/hr.
$\displaystyle d_2 = 45 - 45t$
you see, after 1 hr, $\displaystyle d_2 = 45 - 45(1) = 0$ km from the origin. it made it back.