maths help (1 Viewer)

[authenticity]

Highway A and Highway B intersect perpendicularly. A car on Highway A is 80km from the intersection and is traveling towards it at 50km/h. A car on Highway B is 70km from the intersection and is traveling towards it at 45km/h.
A) find an expression for the square of the distance (x) between the two cars in this manner for h hours.
B) If the cars can continue through the intersection and remain on the same highways, in how many minutes will the distance between them be minimum?

Thanks.

 

[aliqanber]

A) v=d/t ==> d=vt := vh

distance x1 (km) of car on Highway A as function of time (in hours):
x1 = initial distance - change in position = 80km - vh (-vh since the car is getting closer to the intersection)
So, x1 = 80 - 50h

Similarly;
distance x2 (km) of car on Highway B as function of time (in hours):
x2 = initial distance - change in position = 70km - vt
So, x2 = 70 - 45t

By Pythagoras theorem, the distance between both cars as a function of time is;
x = sqrt( x1^2 + x2^2) = sqrt( (80 - 50t)^2 + (70 - 45t)^2 )
x = sqrt( 6400 - 8000t + 2500t^2 + 4900 - 6300t + 2025t^2 )
x = sqrt( 4525t^2 - 14300t + 11300 )
x^2 = 4525t^2 - 14300t + 11300
x^2 = 25 * ( 181t^2 - 572t + 452 ) --------- Answer

B) differentiate x and find critical points and then find the minimum by differentiating x again (Assuming you have done some calculus).