A spy in a speed boat is being chased down a
river by government officials in a faster craft.
Just as the officials’ boat pulls up next to the
spy’s boat, both boats reach the edge of a 4.6
m waterfall. The spy’s speed is 16 m/s and
the officials’ speed is 27 m/s.
How far apart will the two vessels be when
they land below the waterfall? The acceleration of gravity is 9.81 m/s squared
time to fall= sqrt(h/2g)=.48 sec check that
distance difference=speeddifference*timeToFall
= (27-16)*.48= about five meters
To find the distance apart when the two vessels land below the waterfall, we can calculate the time it takes for each boat to reach the edge of the waterfall and then use that time to find the distance they would have traveled.
Let's start with the spy's boat:
The spy's initial speed is 16 m/s, and the acceleration due to gravity is -9.81 m/s² (negative because it acts in the opposite direction). The time it takes for the spy's boat to reach the edge of the waterfall can be found using the equation:
s = ut + 0.5at²
Where:
s = distance covered (unknown)
u = initial velocity (16 m/s)
t = time taken (unknown)
a = acceleration due to gravity (-9.81 m/s²)
Since the final velocity is 0 m/s when the boat reaches the edge of the waterfall, we can rewrite the equation as:
0 = 16t - 0.5 * 9.81 * t²
Simplifying the equation and solving for t:
4.9t² = 16t
Dividing both sides by t:
4.9t = 16
t ≈ 16 / 4.9
t ≈ 3.27 seconds
Now, let's calculate the distance traveled by the spy's boat in this time:
s = ut + 0.5at²
s = 16 * 3.27 + 0.5 * (-9.81) * (3.27)²
s ≈ 52.32 - 50.45
s ≈ 1.87 meters
Therefore, the spy's boat will be approximately 1.87 meters away from the edge of the waterfall when it lands below.
Now let's calculate the time and distance for the officials' boat:
The officials' initial speed is 27 m/s. Using the same equation:
0 = 27t - 0.5 * 9.81 * t²
Simplifying and solving for t:
4.9t² = 27t
Dividing both sides by t:
4.9t = 27
t ≈ 27 / 4.9
t ≈ 5.51 seconds
Calculating the distance:
s = ut + 0.5at²
s = 27 * 5.51 + 0.5 * (-9.81) * (5.51)²
s ≈ 148.77 - 73.51
s ≈ 75.26 meters
Therefore, the officials' boat will be approximately 75.26 meters away from the edge of the waterfall when it lands below.
The distance between the two vessels when they land below the waterfall is the difference between their distances from the edge of the waterfall:
Distance apart = 75.26 - 1.87
Distance apart ≈ 73.39 meters
Therefore, the two vessels will be approximately 73.39 meters apart when they land below the waterfall.
To determine how far apart the two vessels will be when they land below the waterfall, we can use the equation of motion. First, let's find out how long it will take for both boats to reach the waterfall.
For the spy's boat:
Initial velocity, u1 = 16 m/s
Acceleration, a = 9.81 m/s^2 (acceleration due to gravity)
Distance, s1 = ?
Using the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time, we can solve for t1:
s1 = u1t1 + (1/2)at1^2
Since the spy's boat is moving horizontally, the vertical distance traveled doesn't affect the horizontal distance. At the moment the boats reach the edge of the waterfall, their horizontal distances will be equal, so we can ignore the vertical motion.
Since the officials' boat is faster, it will take less time for them to reach the waterfall. Let's call this time t2.
Now, let's find t1 and t2:
For the spy's boat:
s1 = 16t1
For the officials' boat:
s2 = 27t2
Since s1 = s2 at the moment they reach the waterfall:
16t1 = 27t2
Now, let's solve for t1 or t2:
t1 = (27/16)t2
Now, let's substitute this value of t1 into the equation for s1:
s1 = 16[(27/16)t2]
s1 = 27t2
Now, we know that s1 represents the horizontal distance traveled by the spy's boat when they reach the waterfall. To find out how far apart the two vessels will be when they land below the waterfall, we need to subtract the distance traveled by the officials' boat from the distance of the spy's boat:
distance apart = s1 - s2
distance apart = 27t2 - (27t2)
distance apart = 0
The two vessels will be 0 meters apart after landing below the waterfall.