here is a calculus problem..
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who wants to solve it?
A man walks north from a point P at 4m/s. 5 minutes later a woman starts to walk south from a point 500 ft east of P at 5m/s.
Now can someone calculate the rate at which the distance between them increases 10 mins after the woman starts to walk?
I would prefer if you showed me the working. -
x = distance, v = speed, t = time.
x = vt
Differentiate w.r.t to t, then:
dx/dt = dv/dt + v
But since dv/dt = a = acceleration = zero at constant speed, hence
dx/dt = v
Hence, the rate of change of distance is simply the difference between the speed of both person, that is,
dx/dt = 5 - (- 4) = 9m/s (since the lady is walking in the opposite direction as the man, I'm taking the man's direction as positive)
Don't know correct or not. -
My A Maths is F9, so I don't know.
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1 + 1 = 1Originally posted by valentinoo:I would prefer if you showed me the working.
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why the velocity in metres/second and the distance in feet?Originally posted by valentinoo:who wants to solve it?
A man walks north from a point P at 4m/s. 5 minutes later a woman starts to walk south from a point 500 ft east of P at 5m/s.
Now can someone calculate the rate at which the distance between them increases 10 mins after the woman starts to walk?
I would prefer if you showed me the working. -
dont forget the woman is 500ft east of him.Originally posted by CoolMyth:x = distance, v = speed, t = time.
x = vt
Differentiate w.r.t to t, then:
dx/dt = dv/dt + v
But since dv/dt = a = acceleration = zero at constant speed, hence
dx/dt = v
Hence, the rate of change of distance is simply the difference between the speed of both person, that is,
dx/dt = 5 - (- 4) = 9m/s (since the lady is walking in the opposite direction as the man, I'm taking the man's direction as positive)
Don't know correct or not. -
Yeah, my answer wrong....
Sorry, can't help this time. -
i'll post the solution after i get back from the gym, i just finished this topic few weeks ago.
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is it 9m/s??
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it should be 9m/s.
By the way, how come the man can walk at 4m/s? heheheh..
qing gong issit? very fast leh...4m/s?
the woman even faster, at 5m/s, "qing tian dian shui" or "ling bo wei bu" issit? or a ghost?
anyway, the 500ft is irrelevant to the question. And whether its 5min or 10min or even 100min after the woman starts walking is also irrelvant, since there is NO acceleration, ONLY constant velocity. So the RATE is at a constant 9m/s.
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I got 7.46m/s.Originally posted by valentinoo:is it 9m/s?? -
Oooh... so desu ne..Originally posted by nightzip:it should be 9m/s.
By the way, how come the man can walk at 4m/s? heheheh.. qing gong issit? very fast leh...4m/s?
the woman even faster, at 5m/s, "qing tian dian shui" or "ling bo wei bu" issit? or a ghost?
anyway, the 500ft is irrelevant to the question. And whether its 5min or 10min or even 100min after the woman starts walking is also irrelvant, since there is NO acceleration, ONLY constant velocity. So the RATE is at a constant 9m/s. -
since velocity is a vector, then we must account for the component of the difference of the velocity vectors along the straight line between the two persons.
the angle or deviation is small in this case, about 1.32° at t=10 mins after the woman sets off
so the rate of change of distance between them at t=10 mins is given as the velocity component between the two persons at t=10 mins:
| 4-(-5) | × cos 1.32°
= 8.9976 m/s in the direction of the straight line between them -
where u get 1.32 degree??Originally posted by poon cho tang:since velocity is a vector, then we must account for the component of the difference of the velocity vectors along the straight line between the two persons.
the angle or deviation is small in this case, about 1.32° at t=10 mins after the woman sets off
so the rate of change of distance between them at t=10 mins is given as the velocity component between the two persons at t=10 mins:
| 4-(-5) | × cos 1.32°
= 8.9976 m/s in the direction of the straight line between them -
u can make a drawing of their paths after they have set offOriginally posted by valentinoo:where u get 1.32 degree??
if u draw a line connecting the 2 persons, you will form 2 similar triangles, which have a common angle at their far ends
calculate the distances that the man and woman travelled over the full 15 mins, and u will have the lengths of the base of the 2 triangles
proportionate the 500 ft over the distances travelled and u would have obtained the heights of the 2 triangles
use inv tan to find the angle and it will be about 1.32°