My Notes A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 f
t/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?
2 answers:
Answer:
6.667 ft/s
Step-by-step explanation:
Given
Height of street light = 15ft
Height of man = 6ft
Speed away from pole = 4ft/s
Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.
This forms a similar triangle (see attachment below).
From similar triangles, we have
(y - x)/6 = y/15 ----- Solve equation
15(y - x) = 6 * y
15y - 15x = 6y ---- Collect Like Terms
15y - 6y = 15x
9y = 15x --- divide through by 9
y = 15x/9
y = 5x/3 ---- differentiate with respect to time, t
dy/dt = 5/3 dx/dt
dx/dt represents rate of change of distance per time = 4ft/s
while dy/dt represents rate of movement of his shadow tips
dy/dt = 5/3 * 4
dy/dt = 20/3 = 6.667 ft/s
Answer: 6.67ft/s
Step-by-step explanation: PLEASE SEE PICTURE ATTACHED, IT IS A DIAGRAM THAT HELPS YOU TO ANALYSE THE QUESTION.
STEP 1: DEFINE ALL VARIABLES
The man's shadow distance from the pole (Y)
The man's height=6ft
The height of the pole= 15ft
The man's shadow length (X-Y)
The man's speed = 4ft/s
STEP 2: FIND THE MAN'S SHADOW DISTANCE FROM THE POLE (Y)
((X-Y)÷Y) = 6ft÷15ft
Open up bracket and cross multiply
15(X-Y)=6Y
15X-15Y=6Y
collecting like terms together
15X=9Y
Y= 15X/9
Y= 5X/3
STEP 3: FIND THE SPEED OF THE MAN'S SHADOW
Dy/dt = 4ft/s
Therefore
Dy/dx= (4ft/s) × (5X/3)
20ft/3s = 6.67ft/s
The speed of the shadow a cross the pole is 6.67ft/s
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