A train at a constant 41.0 km/h moves east for 38 min, then in a direction 44.0° east of due north for 25.0 min, and then west for 59.0 min. What are the (a) magnitude (in km/h) and (b) angle (relative to north, with east of north positive and west of north negative) of its average velocity during this trip?
First travel segment is 41*38/60 = 25.97 km east.
Second travel segment is 41*25/60 = 17.08 km, 44 degrees east of north.
That has components of 17.08 sin44 = 11.87 km east and 17.08 cos44 = 12.29 kn north.
The third displacement is 40.32 km weast (-40.32 km west).
Add up the total resultant diplacements north and east. Divide them by 38 + 25 + 59 = 122 minutes = 2.067 hours to get average (resultant) velocity components in km/h.
The angle east of north is arctan Rx/Ry, where Rx and Ry are the resultant displacements east and north.
What exactly is the angle. I got 100.55 degrees and it says it is wrong
To find the average velocity of the train, we need to calculate the total displacement and the total time taken.
Step 1: Calculate the displacement in each leg of the journey.
In the first leg, the train moves east for 38 min at a constant speed of 41.0 km/h. The displacement can be calculated using the formula:
Displacement = Velocity x Time
Displacement = 41.0 km/h x (38 min / 60 min/h)
Displacement = 41.0 km/h x 0.6333 h
Displacement = 26.0167 km east
In the second leg, the train moves in a direction 44.0° east of due north for 25.0 min. The displacement can be calculated using trigonometry.
Displacement = cos(44.0°) x Velocity x Time
Displacement = cos(44.0°) x 41.0 km/h x (25.0 min / 60 min/h)
Displacement = 0.7193 x 41.0 km/h x 0.4167 h
Displacement = 12.5297 km north
In the third leg, the train moves west for 59.0 min. The displacement is purely in the west direction, so it is negative.
Displacement = -41.0 km/h x (59.0 min / 60 min/h)
Displacement = -41.0 km/h x 0.9833 h
Displacement = -40.2513 km west
Step 2: Calculate the total displacement.
The total displacement is the vector sum of the individual displacements. We can add the east and north components and subtract the west component.
Total displacement = East displacement + North displacement + West displacement
Total displacement = 26.0167 km east + 12.5297 km north - 40.2513 km west
Total displacement = -1.705 km
Step 3: Calculate the total time taken.
The total time taken is the sum of the times for each leg.
Total time taken = 38 min + 25 min + 59 min
Total time taken = 122 min = 2.0333 h
Step 4: Calculate the average velocity.
Average velocity = Total displacement / Total time taken
Average velocity = -1.705 km / 2.0333 h
Average velocity = -0.8388 km/h
(a) The magnitude of the average velocity is 0.8388 km/h.
(b) The angle of the average velocity relative to north can be calculated using trigonometry. Since the displacement is negative, the angle will be east of north.
Angle = tan^(-1)(North displacement / East displacement)
Angle = tan^(-1)(12.5297 km / 26.0167 km)
Angle = 25.162° east of north
Therefore, the (a) magnitude of the average velocity is 0.8388 km/h and (b) the angle relative to north is 25.162° east of north.
To find the average velocity of the train, we need to calculate both the magnitude and the angle of its resultant displacement.
Let's break down the train's motion into three separate parts:
1. Eastward Motion:
The train moves east at a constant speed of 41.0 km/h for 38 minutes. To find the displacement in this direction, we can use the formula: displacement = velocity × time.
The eastward displacement can be calculated as follows:
Displacement = (41.0 km/h) × (38/60) h = 26.13 km east.
2. Northward Motion:
Next, the train moves in a direction 44.0° east of due north for 25.0 minutes.
To find the northward displacement, we need to calculate the vertical component of the displacement. We can use the formula: vertical displacement = displacement × sin(angle).
The vertical displacement can be calculated as follows:
Vertical displacement = (41.0 km/h) × (25/60) h × sin(44.0°) = 9.61 km.
3. Westward Motion:
Finally, the train moves west at an unknown speed for 59.0 minutes. In this case, we need to subtract the westward displacement from the eastward displacement to find the net horizontal displacement:
Horizontal displacement = (41.0 km/h) × (59/60) h - 26.13 km = 24.95 km west.
Now, we can find the resultant displacement by combining both the horizontal and vertical displacements using the Pythagorean theorem:
Resultant displacement = √[ (horizontal displacement)² + (vertical displacement)² ]
= √[ (24.95 km)² + (9.61 km)² ]
= 26.96 km.
To find the magnitude of the average velocity, we divide the total displacement by the total time taken during the trip:
Total time = 38 minutes + 25 minutes + 59 minutes = 122 minutes = 2.03 hours.
Magnitude of average velocity = (Total displacement) / (Total time)
= 26.96 km / 2.03 h
≈ 13.28 km/h.
The angle (relative to north) of the average velocity can be found using the inverse tangent function:
Angle = arctan[(vertical displacement) / (horizontal displacement)]
= arctan(9.61 km / (-24.95 km))
≈ -20.98°.
Hence, the (a) magnitude of the average velocity is approximately 13.28 km/h, and the (b) angle, relative to north, is approximately -20.98°.