Question

A train at a constant 60.0 km/h maves east for 38.0 min, then in a direction 500% east of due north for 26.0 min, and then west for 48.0 min. What are the (a) magnitude and (D) ang (relative to east) of its average velocity duning this trip? (a) Number Units Units

Answer 1

here,

speed of train , v = 60 km/h

t1 = 38 min = 0.633 hr

displacement 1 , s1 = v * t1 i = 60 km/h * 0.633 h i

s1 = 38 km i

t2 = 26 min = 0.433 hr

displacement 2 , s2 = v * t2 * ( cos(50) i +sin(50) j)

s2 = (16.7 i km + 19.9 j km)

t3 = 48 min = 0.8 hr

displacement 3 , s3 = v * t3 (-i)

s3 = 48 km (-i)

the net displacement , s = s1 + s2 + s3

s = 6.7 km i + 19.9 km j

the average velocity , w = s /t

w = (6.7 km i + 19.9 km j) /(1.866 hr)

w = 3.59 rad/s i + 10.64 rad/s j

a)

the magnitude of net angular velocity , |w| = sqrt(3.59^2 + 10.64^2) = 11.23 rad/s

b)

the direction of angular velocity , theta = arctan(10.64 /3.59)

theta = 71.4 degree North of East