A man walks across a bridge at the rate of 5 ft/s as a boat passes directly beneath him at 10 ft/s. If the bridge is 10 ft above the boat, how fast are the man and the boat separating 1 second later?
Well, the man is walking at 5 ft/s, and the boat is passing directly beneath him at 10 ft/s. So, we can say that they are going in opposite directions, which makes sense because our lives wouldn't be interesting without a little drama.
Now, since the bridge is 10 ft above the boat, we can think of it as a race between the man and the boat, with the bridge serving as the finish line. The man starts ahead because he's on the bridge, but the boat is catching up.
After 1 second, the man would have walked 5 ft, while the boat would have traveled 10 ft. So, the separation between the man and the boat would be the difference between their distances traveled, which is 10 ft - 5 ft = 5 ft.
Therefore, 1 second later, the man and the boat would be separating at a rate of 5 ft/s. Although, I wonder if the man will try to bridge the gap between them with a joke!
To solve this problem, we can use the concept of relative motion.
Let's break it down step by step:
Step 1: Determine the man's and boat's velocities relative to each other.
The man's velocity relative to the boat can be calculated by subtracting the boat's velocity from the man's velocity:
Relative velocity of the man with respect to the boat = Man's velocity - Boat's velocity
Relative velocity of the man with respect to the boat = 5 ft/s - (-10 ft/s)
Relative velocity of the man with respect to the boat = 5 ft/s + 10 ft/s
Relative velocity of the man with respect to the boat = 15 ft/s
Step 2: Calculate the rates at which the man and boat are separating.
Since we know that the bridge is 10 ft above the boat, we can consider the separation distance as the sum of the height and distance traveled horizontally by the boat.
Separation distance = height + horizontal distance
Separation distance = 10 ft + (Relative velocity of the man with respect to the boat * Time)
Separation distance = 10 ft + (15 ft/s * 1 s)
Separation distance = 10 ft + 15 ft
Separation distance = 25 ft
Step 3: Determine how fast the man and the boat are separating.
The rate at which the man and boat are separating is simply the separation distance divided by time:
Rate of separation = Separation distance / Time
Rate of separation = 25 ft / 1 s
Rate of separation = 25 ft/s
Therefore, the man and the boat are separating at a rate of 25 ft/s one second later.
To determine how fast the man and the boat are separating, we need to calculate the relative velocity between them.
First, let's determine the velocity of the man relative to the ground. We are given that the man walks across the bridge at a rate of 5 ft/s, so his velocity relative to the ground is 5 ft/s.
Next, let's determine the velocity of the boat relative to the ground. We are given that the boat passes directly beneath the man at a rate of 10 ft/s. Since the boat is moving directly beneath the man, its velocity relative to the ground is also 10 ft/s.
Now, we can calculate the relative velocity between the man and the boat. Since they are moving in the same direction, we subtract their velocities:
Relative velocity = Velocity of boat - Velocity of man
= 10 ft/s - 5 ft/s
= 5 ft/s
Therefore, the man and the boat are separating at a rate of 5 ft/s when 1 second later.
at 1 second d = sqrt(10^2 + 10^2 + 5^2)
= sqrt 225
d^2 = 100 + 25 t^2 + 100 t^2
d^2 = 100 + 250 t^2
2 d dd/dt = 500 t
v = dd/dt = 250 t/d
at t = 1
v = 250/15 = 16.7 ft/s