Please help!!!
A jetliner flies west 2,000 miles from Chicago you San Francisco. It has an average speed of 635 mph relative to the air. What is the velocity of the jetliner relative to the earth of the wind is blowing at 160 mph from the west to the east?
475 mph, east
475 mph, west(my guess)
795 mph, west
795 mph, east
A hot air ballon rises vertically at a constant rate of 8 m/s. A second hot air balloon descends passed the first hot air balloon at rate of 12 m/s relative to the first hot air balloon. What is the velocity of the second hot air ballon relative to the ground?
4 m/s, Up
20 m/s. Up
4 m/s, down
20 m/s, down(my guess)
To calculate the velocity of the jetliner relative to the earth, we need to calculate the net speed by considering both the speed of the jetliner and the speed of the wind.
1. Given:
- Speed of the jetliner relative to the air: 635 mph (west).
- Speed of the wind: 160 mph (east).
2. To find the velocity of the jetliner relative to the earth, we need to subtract the speed of the wind from the speed of the jetliner since they are in opposite directions.
3. Calculating the net speed:
Velocity of the jetliner relative to the earth = Speed of the jetliner - Speed of the wind
= 635 mph (west) - 160 mph (east)
4. Subtraction:
Velocity of the jetliner relative to the earth = 635 mph - 160 mph
= 475 mph (west)
So, the correct answer is 475 mph, west.
Now let's move on to the second question regarding the velocity of the second hot air balloon relative to the ground.
1. Given:
- Vertical rate of ascent of the first hot air balloon: 8 m/s (upward).
- Vertical rate of descent of the second hot air balloon relative to the first hot air balloon: 12 m/s (downward).
2. To find the velocity of the second hot air balloon relative to the ground, we need to add the rate of ascent of the first balloon and the rate of descent of the second balloon.
3. Calculating the net velocity:
Velocity of the second hot air balloon relative to the ground = Rate of ascent of the first balloon + Rate of descent of the second balloon
= 8 m/s (upward) + 12 m/s (downward)
4. Addition:
Velocity of the second hot air balloon relative to the ground = 8 m/s + (-12 m/s)
= -4 m/s (downward)
So, the correct answer is 4 m/s, down.
To find the velocity of the jetliner relative to the earth, taking into account the wind speed, we can use vector addition.
The velocity of the jetliner relative to the air is given as 635 mph, and the wind is blowing at 160 mph from the west to the east. Since the wind is blowing in the opposite direction of the jetliner's movement, we subtract the wind velocity from the jetliner's velocity to get the velocity of the jetliner relative to the earth.
635 mph (west) - 160 mph (east) = 475 mph (west)
Therefore, the velocity of the jetliner relative to the earth is 475 mph, west.
For the second question about the hot air balloons:
The first hot air balloon rises vertically at a constant rate of 8 m/s. The second hot air balloon descends relative to the first balloon at a rate of 12 m/s.
To find the velocity of the second hot air balloon relative to the ground, we can use vector addition.
The velocity of the second hot air balloon relative to the first balloon is 12 m/s (down). However, we want to find the velocity of the second hot air balloon relative to the ground.
Since the first balloon is rising at 8 m/s, we subtract the velocity of the first balloon from the velocity of the second balloon to get the velocity of the second balloon relative to the ground.
12 m/s (down) - 8 m/s (up) = 4 m/s (down)
Therefore, the velocity of the second hot air balloon relative to the ground is 4 m/s, down.