Question
Ian's house is located 20km north of Aida's house. At 9am, one saturday, Ian leaves his house and jogs south at 8km/hour. At the same time Aida leaves her house and jogs east at 6km/h. When are Ian and Aida closest together given that they both run for 2.5 hours? The answer is t=1.6 hours @ 10:36 am
PLEASE SHOW sTEP bY STEP!! i don't understand how to do this
PLEASE SHOW sTEP bY STEP!! i don't understand how to do this
Answers
draw a sketch of the situation,
at a time of t hours after 9:00 am
let A1 be the position of Aida
let I1 be the position of Ian
Join A1 and I1, to get a right-angled triangle
let D be the distance between them
D^2 = (6t)^2 + (20-8t)^2
= 36t^2 + 400 - 320t + 64t^2
= 100t^2 - 320t + 400
2D dD/dt = 200t - 320
dD/dt = (200t-320)/(2D)
= 0 for a minimum of D
200t = 320
t = 1.6
So they are closest after 1.6 hours or 1 hour and 36 minutes past 9:00 am
making it 10:36 am
This is a standard question for this topic of Calculus.
In my opinion, you should really know how to do this question
at a time of t hours after 9:00 am
let A1 be the position of Aida
let I1 be the position of Ian
Join A1 and I1, to get a right-angled triangle
let D be the distance between them
D^2 = (6t)^2 + (20-8t)^2
= 36t^2 + 400 - 320t + 64t^2
= 100t^2 - 320t + 400
2D dD/dt = 200t - 320
dD/dt = (200t-320)/(2D)
= 0 for a minimum of D
200t = 320
t = 1.6
So they are closest after 1.6 hours or 1 hour and 36 minutes past 9:00 am
making it 10:36 am
This is a standard question for this topic of Calculus.
In my opinion, you should really know how to do this question