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# Answer to Question #21618 in Mechanics | Relativity for Abdulrahman Hussein

Question #21618
a train moving at an essentially constant speed of 60 miles/hr moves eastward for 40 min,then in a direction 45 degree east of north for 20 min,and finally westward for 50 min. what is the average velocity of the train during this run?
Expert's answer
Let the train initially be at the origin of the coordinate plane xOy, with positive y-direction to the north and positive x-direction to the east.
The way is divided into three parts.
The times of the parts are: t1 = 40 min = 2/3 hr; t2 = 20 min = 1/3 hr; t3 = 50 min = 5/6 hr.
Angles of the motion (counterclockwise from the positive x - direction): a1 = 0; a2 = 45 degree; a3 = 180 degree.
The distances moved during each part: S1 = v*t1 = 60 miles/hr * 2/3 hr = 40 miles; S2 = v*t2 = 60 miles/hr * 1/3 hr = 20 miles;
S3 = v*t3 = 60 miles/hr * 5/6 hr = 50 miles.
The distances moved along each axis during each part: x1 = s1*cos(a1) = 40*cos(0) = 40 miles; y1 = s1*sin(a1) = 40*sin(0) = 0;
x2 = s2*cos(a2) = 20*cos(45) = 14.14 miles; y2 = s2*sin(a2) = 20*sin(45) = 14.14 miles;
x3 = s3*cos(a3) = 50*cos(180) = -50 miles; y3 = s3*sin(a3) = 50*sin(180) = 0;
The total distance moved along the axes:
x = x1+x2+x3 = 40 + 14.14 - 50 = 4.14 miles
y = y1+y2+y3 = 0 + 14.14 + 0 = 14.14 miles
The total distance travelled:
s = sqrt(x*x + y*y) = sqrt(4.14^2 + 14.14^2) = 14.74 miles
The total time spent: t = t1+t2+t3 = 40 min + 20 min + 50 min = 110 min = 11/6 hr

So, the average velocity v = s/t = 14.74 miles / (11/6 hr) = 8.04 miles/hr
The average angle of motion: alpha = asin(y/s) = asin(14.14 / 14.74) = 73.7 degrees (north of east)

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