Asked by Jasmine
A man starts walking north at 2ft/s from a point P. Twenty-five seconds later a woman starts walking south at 3ft/s from a point 400ft due east of P. At what rate are the people moving apart 50 seconds after the woman starts walking? Do not include units in your answer.
Answers
Answered by
Damon
Man starts north at t = 0 from origin
Woman starts south at t = 25 from (400,0)
where are they at t = 75?
man 2*75 = 150 so at (0,150)
woman 3*50 = 150 so at (400,-150)
distance apart is hypotenuse of triangle with legs 300 and 400 so 500 ft
How fast is this changing?
The north south difference is changing at 2+3 = 5 ft/s
The East west distance is not changing at all
start new t now at 0
N-S distance = 300 + 5 t
E-W distance = 400
h is hypotenuse, the distance
h^2 = 400^2 + (300+5t)^2
2 h dh = 0 + 2 (300+5t)(5)dt
h dh = 5(300+5t) dt
at h = 500 and t = 0 this is
500 dh = 5(300) dt
dh/dt = .01(300) = 3
Woman starts south at t = 25 from (400,0)
where are they at t = 75?
man 2*75 = 150 so at (0,150)
woman 3*50 = 150 so at (400,-150)
distance apart is hypotenuse of triangle with legs 300 and 400 so 500 ft
How fast is this changing?
The north south difference is changing at 2+3 = 5 ft/s
The East west distance is not changing at all
start new t now at 0
N-S distance = 300 + 5 t
E-W distance = 400
h is hypotenuse, the distance
h^2 = 400^2 + (300+5t)^2
2 h dh = 0 + 2 (300+5t)(5)dt
h dh = 5(300+5t) dt
at h = 500 and t = 0 this is
500 dh = 5(300) dt
dh/dt = .01(300) = 3
Answered by
Reiny
let the times passed since the woman started walking be t seconds
make a diagram
distance covered by man = 50 + 2t
distance covered by woman = 3t
let the distance between them be D
I see a right-angled triangle with hypotenuse D,
the vertical line as (50+2t+3t)
and the horizontal as 400
D^2 = (5t+50)^2 + 400^2
2D dD/dt = 2(5t+50)(5) + 0
dD/dt = 5(5t+50)/D
when t = 50,
D^2 = 300^2+400^2
D = 500
dD/dt = 5(300)500 = 3
make a diagram
distance covered by man = 50 + 2t
distance covered by woman = 3t
let the distance between them be D
I see a right-angled triangle with hypotenuse D,
the vertical line as (50+2t+3t)
and the horizontal as 400
D^2 = (5t+50)^2 + 400^2
2D dD/dt = 2(5t+50)(5) + 0
dD/dt = 5(5t+50)/D
when t = 50,
D^2 = 300^2+400^2
D = 500
dD/dt = 5(300)500 = 3
Answered by
Jasmine
thank you very much
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