An antelope is at a distance of 20.0 m from a converging lens of focal length 32.0 cm. The lens forms an image of the animal. If the antelope runs towards the lens at a speed of 5.50 m/s, what is the speed with which the image moves?
1/q= 1/32cm-1/2000cm
q=32.5203252
dq/dt=-(32.5203252/2000)^2 *5.50m/s=-1.45mm/s(the sign does not mater), i don't get why this is wrong
You have the equations correct but the answer is wrong by a multiple of ten. Do the problem again, but all in meters and change it to mm in the end.
The equation you used to calculate the image distance (q) is correct: 1/q = 1/f - 1/p, where f is the focal length and p is the object distance.
However, to find the speed with which the image moves, you need to differentiate this equation with respect to time (t) and use the chain rule. The equation becomes:
(dq/dt) = -[(1/f) - (1/p)] * (dp/dt)
In this case, dp/dt represents the speed at which the object (the antelope) is moving towards the lens.
Given:
f = 32 cm (0.32 m)
p = 20.0 m
dp/dt = 5.50 m/s
Substituting these values into the equation:
(dq/dt) = -[(1/0.32) - (1/20.0)] * 5.50
Simplifying this expression:
(dq/dt) = -[(1/0.32) - (0.05)] * 5.50
(dq/dt) = -[(3.125 - 0.05)] * 5.50
(dq/dt) = -[3.075] * 5.50
(dq/dt) = -16.91 mm/s
Therefore, the correct speed with which the image moves is -16.91 mm/s (note that the negative sign indicates that the image is moving towards the lens).
To understand why the answer you got is wrong, let's go through the calculation together step by step.
The formula you used for the lens is correct:
1/f = 1/p - 1/q
where:
f is the focal length of the lens (in meters)
p is the object distance (in meters)
q is the image distance (in meters)
Given:
f = 32.0 cm = 0.32 m (converted to meters)
p = 20.0 m
Substituting these values into the formula:
1/0.32 = 1/20 - 1/q
Simplifying:
3.125 = 0.05 - 1/q
Rearranging the equation to solve for q:
1/q = 0.05 - 3.125
1/q = -3.075
Taking the reciprocal of both sides:
q = -1/3.075
Now, let's find the rate at which the image moves (dq/dt) as the antelope runs towards the lens.
dq/dt = -(1/q^2) * dp/dt
dp/dt is the rate at which the object distance (p) is changing, which is given as the antelope's speed: dp/dt = -5.50 m/s (negative because the antelope is moving towards the lens)
Substituting the values into the equation:
dq/dt = -((-1/3.075)^2) * (-5.50 m/s)
= -1.45 mm/s
The negative sign simply indicates that the image moves in the opposite direction as the object. In this case, since the antelope is running towards the lens, the image will move away from the lens.
So, your calculation of -1.45 mm/s for the image speed is correct.