Rate..................m/sec
Give your answer upto 3 decimal places.)
Give your answer upto 3 decimal places.)
-
Find the rate at which his?
-- hair grows?
-- bowling score improves?
-- car depreciates in value?
OK, drawing on experience, I've never seen a question that starts this way, that doesn't end in something to the effect,
(a) -- shadow moves?
or
(b) -- shadow grows?
If you idealize:
the lamp as a point source, S,
his head a a point object, P, at a height, h = 1.2 m
- above an exactly level ground plane, G,
then all the possible positions of P are in a plane, A, at height h above G. Call the origin in each plane (A and G), the point at the foot of the perpendicular from S onto each plane. This means that the rays from S form shadows in G of each point of A that are just a constant multiple of their distances from the origin; plane G is just a magnification of plane A. You can see this by drawing a ray from S through A onto G and identifying similar right triangles (the perpendicular from S forming one side). The magnifying ratio is just the ratio:
r = (distance from S to G)/(distance from S to A) = 5 m / (5 m - 1.2 m) = 25/19
(a) So any motion of P at speed v in plane A will cast a shadow that moves at speed
u = r*v = 25v/19
In your example, v = 3.2 m/s, so
u = 80/19 m/s = 4.21... m/s
(b) That answers (a). For (b), you just subtract the speed he's moving:
w = u - v = (4.21... - 3.2) m/s = 1.01... m/s
-- hair grows?
-- bowling score improves?
-- car depreciates in value?
OK, drawing on experience, I've never seen a question that starts this way, that doesn't end in something to the effect,
(a) -- shadow moves?
or
(b) -- shadow grows?
If you idealize:
the lamp as a point source, S,
his head a a point object, P, at a height, h = 1.2 m
- above an exactly level ground plane, G,
then all the possible positions of P are in a plane, A, at height h above G. Call the origin in each plane (A and G), the point at the foot of the perpendicular from S onto each plane. This means that the rays from S form shadows in G of each point of A that are just a constant multiple of their distances from the origin; plane G is just a magnification of plane A. You can see this by drawing a ray from S through A onto G and identifying similar right triangles (the perpendicular from S forming one side). The magnifying ratio is just the ratio:
r = (distance from S to G)/(distance from S to A) = 5 m / (5 m - 1.2 m) = 25/19
(a) So any motion of P at speed v in plane A will cast a shadow that moves at speed
u = r*v = 25v/19
In your example, v = 3.2 m/s, so
u = 80/19 m/s = 4.21... m/s
(b) That answers (a). For (b), you just subtract the speed he's moving:
w = u - v = (4.21... - 3.2) m/s = 1.01... m/s