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?
1
Expert's answer
2013-01-14T09:32:25-0500
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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