Two trains are passing against each other.?
Two trains are passing against each other.
train A is going north with velocity of 47 km/hour.
train B is going south with velocity of 94 km/hour.
a passenger in train A see the smoke that emitted from his train chimney making
an angle of 60 degrees from the west to the south.
a passenger in train B see the smoke that emitted from his train chimney making
an angle of 30 degrees from the south to the north.
a) What is the velocity of the wind – in respect to the ground?
the value of the velocity (of the wind) , and the direction (angle) of the velocity of
the wind (how many degrees from the west to the south)
b) what is the velocity of the smoke of train A , in respect to viewer in train A?
c) what is the velocity of the smoke of train B, in respect to viewer in train B ?
To solve this problem, we can break it down into three parts: finding the velocity of the wind, the velocity of the smoke in train A, and the velocity of the smoke in train B.
a) To find the velocity of the wind, we can use the concept of vector addition. Let's assume the wind velocity (Vw) is in the direction from west to south. We need to find the magnitude and direction of the wind velocity.
Let's represent the velocity of train A as Va = 47 km/hour (north) and the velocity of train B as Vb = 94 km/hour (south). We can consider them as vectors.
The angle between the smoke emitted by train A and the wind direction (Vw) is 60 degrees from the west to the south. The angle between the smoke emitted by train B and the wind direction (Vw) is 30 degrees from the south to the north.
Using vector addition, the resultant velocity of the smoke emitted by train A with respect to the ground (VsA) can be represented as:
VsA = Va + Vw
Similarly, the resultant velocity of the smoke emitted by train B with respect to the ground (VsB) can be represented as:
VsB = Vb + Vw
Since the trains are passing against each other, the resultant ground velocity of train B is taken as negative (-Vb).
Now, let's solve for the magnitude and direction:
For train A:
VsA = Va + Vw
VsA = 47 km/hour (north) + Vw (from west to south)
For train B:
VsB = -Vb + Vw
VsB = -94 km/hour (south) + Vw (from west to south)
Since the velocities are given in km/hour, we can solve for the wind velocity (Vw) in km/hour.
b) To find the velocity of the smoke of train A with respect to the viewer in train A, we need to subtract the viewer's frame of reference. Since the viewer is in train A, the velocity of the viewer (VvA) is taken as zero.
The velocity of the smoke of train A with respect to the viewer in train A (VsAvA) can be represented as:
VsAvA = VsA - VvA
VsAvA = VsA - 0
VsAvA = VsA
c) Similarly, to find the velocity of the smoke of train B with respect to the viewer in train B, we need to subtract the viewer's frame of reference. Since the viewer is in train B, the velocity of the viewer (VvB) is taken as zero.
The velocity of the smoke of train B with respect to the viewer in train B (VsBvB) can be represented as:
VsBvB = VsB - VvB
VsBvB = VsB - 0
VsBvB = VsB
To summarize:
a) The velocity of the wind with respect to the ground is Vw (magnitude) in the direction from west to south.
b) The velocity of the smoke emitted by train A with respect to the viewer in train A is VsA (magnitude and direction).
c) The velocity of the smoke emitted by train B with respect to the viewer in train B is VsB (magnitude and direction).