1. A yacht is tacking into the wind on a zigzag path. On the �rst leg of the course, the yacht has a
displacement of 12 km at 840 north of east. After a second leg has been completed, the yacht's resultant
displacement is 10 miles at 230 west of south. Determine the magnitude and direction of the second leg
of the course.
2. A ferry boat crosses a 550-m wide river from the east to the west shore in 1 min and 9 s. (a) What is the
boat's average velocity? (b) The boat makes the return trip 500 m in 58 s. What is the boat's average
velocity for the return trip? (c) What is the boat's average speed for the entire round trip? What is the
boat's average velocity for the entire round-trip?
3. Suppose a gazelle is capable of accelerating from rest to its top speed of 25 m/s in a distance of 50 m
and, after attaining its speed, the animal can maintain it for an extended period of time. In addition,
suppose a cheetah can accelerate from rest to its top speed of 30 m/s in a distance of 60m, but then
can maintain this speed for only 4.0 s before it must give up the chase. How close to a gazelle must a
cheetah be before it can launch a successful attack?
4. A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above.
The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were
the keys thrown? (b) What was the velocity of the keys just before they were caught?
Your umpteen posts have been removed.
Once you write up YOUR THOUGHTS, please re-post, and someone here will be happy to comment.
1. To determine the magnitude and direction of the second leg of the course, we can use vector addition.
First, let's convert the displacements to Cartesian coordinates. For the first leg, the displacement is 12 km at 840 north of east. This can be broken down into its x and y components:
- x component = 12 km * cos(840) = 12 km * cos(60°) = 6 km
- y component = 12 km * sin(840) = 12 km * sin(60°) = 10.39 km
For the second leg, the displacement is 10 miles at 230 west of south. Converting miles to kilometers, we have:
- displacement = 10 miles * 1.609 km/mile = 16.09 km
- x component = 16.09 km * cos(230) = 16.09 km * cos(50°) = 10.37 km
- y component = -16.09 km * sin(230) = -16.09 km * sin(50°) = -12.22 km
Now, to find the second leg's magnitude and direction, we will add the x and y components of both legs:
- x component = 6 km + 10.37 km = 16.37 km
- y component = 10.39 km - 12.22 km = -1.83 km
To find the magnitude, we use the Pythagorean theorem:
magnitude = sqrt((16.37 km)^2 + (-1.83 km)^2) = sqrt(267.8589 km^2) = 16.37 km
To find the direction, we use the inverse tangent function:
direction = atan(-1.83 km / 16.37 km) = -6.39°
Therefore, the magnitude of the second leg of the course is approximately 16.37 km, and it is directed at an angle of -6.39° relative to the positive x-axis.
2. (a) To find the boat's average velocity, we divide the total displacement by the total time taken. In this case, the displacement is 550 m from east to west, and the time taken is 1 min 9 s = 69 s.
average velocity = 550 m / 69 s = 7.97 m/s from east to west.
(b) For the return trip, the displacement is 500 m from west to east, and the time taken is 58 s.
average velocity = 500 m / 58 s = 8.62 m/s from west to east.
(c) To find the boat's average speed for the entire round trip, we divide the total distance traveled by the total time taken. The total distance traveled is 550 m + 500 m = 1050 m, and the total time taken is 69 s + 58 s = 127 s.
average speed = 1050 m / 127 s = 8.27 m/s.
To find the boat's average velocity for the entire round trip, we look at the direction of the displacement. Since the displacement is zero (east to west and then back to east), the boat's average velocity for the entire round trip is zero.
3. To determine how close a cheetah must be to a gazelle for a successful attack, we need to consider the acceleration and top speed of both animals.
The gazelle can accelerate from rest to its top speed of 25 m/s in a distance of 50 m. We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (25 m/s)
u = initial velocity (0 m/s, since it starts from rest)
a = acceleration
s = distance (50 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s) = (25^2 - 0^2) / (2 * 50) = 12.5 m/s^2
The cheetah can accelerate from rest to its top speed of 30 m/s in a distance of 60 m. Using the same equation:
a = (v^2 - u^2) / (2s) = (30^2 - 0^2) / (2 * 60) = 7.5 m/s^2
Now, let's consider that the cheetah starts chasing when both animals are at rest (initial velocity is 0 m/s for the gazelle and cheetah). The time taken for the cheetah to reach a distance s is given by:
t = (v - u) / a = (25 - 0) / 12.5 = 2 s
Since the cheetah can maintain its top speed for only 4.0 s, we need to find how far it can travel during this time. Using the equation of motion:
s = ut + (1/2) a t^2 = 0 * 4 + (1/2) * 7.5 * (4^2) = 30 m
Therefore, the cheetah must be within 30 m of the gazelle before it can launch a successful attack.
4. (a) To find the initial velocity at which the keys were thrown upward, we can use the equation of motion:
s = ut + (1/2)at^2
The final displacement (s) is the height of the window, 4.00 m.
The time taken (t) is the time for the keys to reach the sister's hand, which is 1.50 s.
The acceleration (a) is due to gravity and is -9.8 m/s^2 (since it acts downward).
Rearranging the equation, we get:
u = (s - (1/2)at^2) / t = (4.00 - (1/2)(-9.8)(1.50^2)) / 1.50 = 8.72 m/s
Therefore, the initial velocity at which the keys were thrown upward is approximately 8.72 m/s.
(b) To find the velocity of the keys just before they were caught, we need to consider the effect of gravity. The velocity at any point during the motion can be found using the equation:
v = u + at
Given that the final velocity (v) is 0 m/s (since the keys reach their highest point and then start falling), the acceleration (a) is still -9.8 m/s^2, and the time (t) is 1.50 s.
Rearranging the equation, we get:
0 = u - 9.8 * 1.50
u = 14.7 m/s
Therefore, the velocity of the keys just before they were caught is approximately 14.7 m/s.