A train at a constant 41.0 km/h moves east for 28 min, then in a direction 53.0° east of due north for 15.0 min, and then west for 64.0 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?
41km/h[0o], 28min.
41km/h[37o], 15min.
41km/h[180o], 64min
a. d1=41km/h[0o] * (28/60)h=19.13km[0o]
d2=41km/h[37o] * (15/60)h=10.25km[37o]=
8.19 + i6.17
d3=41km/h[180o] * (64/60)h=43.73km[180o]
V=(d1+d2+d3)/(t1+t2+t3)=(19.13+8.19+i6.17-43.73)/(28+15+64) =
(-16.41 + i6.17)/107=17.53[159.4o]/107 =
0.16384km/min = 9.83 km/h.
b. Tan Ar = Y/X = 6.17/-16.41 = -0.37599
Ar = -20.61 = Reference angle.
A = -20.6 + 180 = 159.4o, CCW. = 69.4o
West of North.
To find the average velocity of the train, we need to calculate the total displacement and total time taken during the trip.
Let's break down the train's motion into three parts:
1. Eastward motion for 28 min at a constant speed of 41.0 km/h.
2. Motion in a direction 53.0° east of due north for 15.0 min.
3. Westward motion for 64.0 min.
For the first part, the displacement (distance) covered is given by:
displacement = speed * time = 41.0 km/h * (28 min / 60 min/h)
Next, we need to find the northward and eastward components of the displacement for the second part. We can use trigonometry to do this.
northward displacement = speed * time * sin(angle) = 41.0 km/h * (15 min / 60 min/h) * sin(53.0°)
eastward displacement = speed * time * cos(angle) = 41.0 km/h * (15 min / 60 min/h) * cos(53.0°)
Finally, for the third part, the displacement (distance) covered is given by:
displacement = speed * time = 41.0 km/h * (64 min / 60 min/h)
To find the total displacement, we need to add up all the displacements (taking into account direction). We can combine the eastward and westward displacements:
total eastward displacement = eastward displacement - westward displacement
total displacement = sqrt((displacement)^2 + (total eastward displacement)^2)
The total time taken is the sum of the individual times:
total time = 28 min + 15 min + 64 min
Now we have the total displacement and total time, we can find the average velocity using the formula:
average velocity = total displacement / total time
Finally, to find the magnitude of the average velocity, we take the absolute value of the average velocity:
magnitude of average velocity = |average velocity|
To find the angle of the average velocity relative to north, we can use inverse tangent (arctan) to find the angle between the northward component of the displacement and the displacement vector:
angle = arctan((total eastward displacement) / (displacement))
Now, let's plug in the given values into the formulas to find the answers:
(a) Magnitude of the average velocity:
- Calculate the total displacement using the given time intervals and speeds.
- Calculate the total time taken for the trip by adding up the given time intervals.
- Plug the values into the average velocity formula: average velocity = total displacement / total time.
- Take the absolute value of the average velocity to find its magnitude.
(b) Angle of the average velocity:
- Calculate the eastward and northward displacements for the second part using the given speed, time, and angle.
- Calculate the total eastward displacement by subtracting the westward displacement from the eastward displacement.
- Calculate the total displacement using the first and third parts' displacements and the total eastward displacement.
- Plug the values into the angle formula: angle = arctan((total eastward displacement) / (displacement)).
- The resulting angle will be the angle of the average velocity relative to north, with east of north positive and west of north negative.