physics help
whats up?
its a question on 1 d motion
put the question out.......
wel a man walks with a uniform velocity v towards a lantern of height H from ground
his height is h
find the velocity of the tip of his shadow
when i checked the answer key it is given V*(H-h/h) but i dont knw jow it is got
Nice question. Let the man be distance D from the lamp and the length of his shadow L. Then H/(D+L) = h/d => DH = h(d+L) => L = (H-h)/h . D Now the only thing changing on the RHS is D. We know D is change at a rate of v so rate of change of L is (H-h)/h . v
(formally, differentiate wrt time L = (H-h)/h . D )
helppppppppppp
What step don't you follow? Draw a diagram certainly to help.
wait dont go away let me read
and draw
post a comment when you're ready and I'll come back.
i am confused which is L and in fig
can u show
..........
.................u thr?
hw=ey there is no small d in fig
d = D. I'm using D
H/(D+L) = h/D => DH = h(D+L) => L = (H-h)/h . D
L = (H-h)/h . D (H-h)/h . i didnt get this
??
DH = h(D+L) => DH = Dh + Lh => Lh = DH - Dh => Lh = D(H-h) => L = (H-h)/h . D
(H-h)/h . v that step
Now differentiate both sides dL/dt = (H-h)/h . dD/dt = (H-h)/h . v
wel how we know that this is vlocity of the shadow??
at tip???????/
because L is the length of the shadow so its derivative is its velocity.
but only with respect to the man.
d is length of shadow ryt?
Yes, by defintion and look at the diagram. with respect to the lamp, the velocity of the the tip of the shadow is dL/dt + v = (H-h)/h v + v = H/h . v
pls show the steps of that v w r to lamp
some more
because the man walking towards the lamp at velocity v
Suppose you're walking a dog. You are walking at speed v. And he walking towards you RELATIVE TO YOU at a speed w. Then the dog is walking at speed of v + w
if w = 0 then he is walking at a speed v and hence he doesn't get any closer or farther away
confused :(
Imagine walking a dog. You're walking at speed v. At what speed does he have to walk so he is not closer or farther away from you?
v
yes and what is his velocity relative to you?
0
Now imagine he is walking with velocity of w relative to you. What is his absolute velocity if you are walking with a velocity of v?
w+v
Right. So that is what is going here with the tip of the shadow. The shadow length D was defined relative to the position of the man.
and we found dD/dt, the velocity of the tip of the shadow relative to the man.
The other way to think about this is the position of the tip of the shadow. That position is L + D
k
hence the velocity of the tip of the shadow is \[\frac{d \ }{dt} (L + D) = \frac{dL}{dt} + \frac{dD}{dt} = v + \frac{H-h}{h} v = \frac{H}{h}v\]
oh k bt in othr case y we dont find relative velocity ??
dD/dt is the velocity relative to the man.
oh i got it as lamp was at rest we dont consider relative velocity ryt??
I think we can assume the lamp hasn't got legs, yes
again dD/dt is absolute velocity??if it was v with respect to man it shuld be vi H-h/h *v
v-(H-h)/h*v
"a man walks with a uniform velocity v towards a lantern of height H from ground"
Are the man and the tip of the shadow moving in opposite directions? No. So it's addition, not subtraction.
y?????
va- vb when same direction
Why are the man and the tip of the shadow moving in the same direction?
NO. Go back back to our dog example.
in dog example absolute value of dog velocity was w+v
but rel is va-vb
yes. added. that's what we're doing here.
but rel is va-vb we are finding relative here
so it's dD/dt
which is the relative veloiccity of the shadow tip
vs-vman
relative to the man that is, because D itself is measured relative to the man.
oh k
i hav another o ne question
post it separately
:)
physicsss
Join our real-time social learning platform and learn together with your friends!