# Average velocity of train

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 $\frac{\Delta r}{\Delta t}$. I am starting at the origin, and have three vectors a , b , and c .

$a = 40i$
$b = (10\sqrt{2}) i + (10\sqrt{2}) j$
$c = -50i$

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

Thanks

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Mentor
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.

Yeah I converted 60 miles/hr to 1 mile/min. So your saying that the average velocity is just $(10\sqrt{2} - 10)^{2}) + (10\sqrt{2})^{2})$? The direction would be $\tan^{-1}(\frac{10\sqrt{2}-10}{10\sqrt{2}})$ so $\theta = 16.3$ degrees north of east?

Thanks

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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 $R_x i + R_y j$, then its magnitude is $\sqrt(R_x^2 + R_y^2)$. To find the magnitude of the average velocity, divide by the time.