What is the average velocity of the train during this run?

In summary, the train traveled eastward for 40 minutes at a constant speed of 60 miles/hr, then moved in a direction 45 degrees east of north for 20 minutes, and finally went westward for 50 minutes. The average velocity can be calculated by finding the resultant vector of the three displacements, dividing it by the total time of 110 minutes, and taking the magnitude of the resulting vector. The average velocity is a vector with a magnitude of (10\sqrt{2} - 10)^{2}) + (10\sqrt{2})^{2}) and a direction of 16.3 degrees north of east.
  • #1
courtrigrad
1,236
2
A train moving at an essentially constant speed of 60 miles/hr moves eastward for 40 min, then in a direction 45 degrees east of north for 20 min, and finally westward for 50 min. What is the average velocity of the train during this run?

So average velocity is [itex] \frac{\Delta r}{\Delta t} [/itex]. I am starting at the origin, and have three vectors a , b , and c .

[itex] a = 40i [/itex]
[itex] b = (10\sqrt{2}) i + (10\sqrt{2}) j [/itex]
[itex] c = -50i [/itex]

So the resultant vector would be [itex] (10\sqrt{2} - 10)i + (10\sqrt{2})j[/itex]. So to find the average velocity would I just do [itex] \frac{(10\sqrt{2} - 10)^{2}) + (10\sqrt{2})^{2})}{110} [/itex]

Thanks
 
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  • #2
The magnitude of the average velocity is the magnitude of that resultant vector. (I think you mistyped your answer.)

Careful about units. Since you have the time in minutes, you are measuring velocity in miles/minute, not the given miles per hour.

The average velocity is a vector: it has both direction (figure out what angle it makes with the x-axis) and magnitude.
 
  • #3
Yeah I converted 60 miles/hr to 1 mile/min. So your saying that the average velocity is just [itex] (10\sqrt{2} - 10)^{2}) + (10\sqrt{2})^{2}) [/itex]? The direction would be [itex] \tan^{-1}(\frac{10\sqrt{2}-10}{10\sqrt{2}}) [/itex] so [itex] \theta = 16.3 [/itex] degrees north of east?

Thanks
 
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  • #4
No, he is saying that the average velocity is a vector- the displacement vector you calaculated divided by the 110 minutes (in miles per minute of course).

Although Doc Al said "it has both direction (figure out what angle it makes with the x-axis) and magnitude." I don't think you really need to do that. One way to give a vector is to give direction (angle) and magnitude, but component form should be perfectly acceptable.
 
  • #5
As Halls said, describing the vector in component form is perfectly acceptable. But if you need the magnitude, calculate it properly. If the resultant displacement vector is [itex]R_x i + R_y j[/itex], then its magnitude is [itex]\sqrt(R_x^2 + R_y^2)[/itex]. To find the magnitude of the average velocity, divide by the time.
 

Related to What is the average velocity of the train during this run?

1. What is the formula for calculating average velocity of a train?

The formula for average velocity is the total displacement of the train divided by the total time it takes to travel that distance.

2. How do you measure the displacement of a train?

The displacement of a train can be measured using a distance measuring tool, such as a ruler or a measuring tape, to determine the distance traveled in a straight line from the starting point to the ending point.

3. Is average velocity the same as average speed?

No, average velocity and average speed are not the same. Average velocity takes into account the direction of the train's motion, while average speed only looks at the magnitude of the train's motion.

4. What units are used to measure average velocity of a train?

The units used to measure average velocity of a train are distance over time, such as meters per second or miles per hour.

5. How does the mass of a train affect its average velocity?

The mass of a train does not directly affect its average velocity. However, it can affect the train's acceleration, which in turn can impact its average velocity. A heavier train may require more force to accelerate, resulting in a lower average velocity compared to a lighter train.

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