A car travelled 60 km west in 1.33 h along a straight line, then returned back east for 40 km in 0.67 h. Its average velocity is:

a) 10 km/h, east.
b) 10 km/h, west.
c) 50 km/h, east.
d) 50 km/h, west.

Option A

Explanation:

Velocity is expressed as distance covered per unit time, with respect to direction. Therefore, v=d/t

Given distance west as 60 km and time as 1.33 then velocity will be

V=60/1.33=45.112781954887 km/h

Rounded off as 45.11 km/h West

Velocity in East will also be given by substituting 40 km for d and 0.67 h for h hence

V=40/0.67=59.701492537313 km/h rounded off as 59.70 km/h East

Taking East as positive then West as negative, the sum of two velocities will be (59.70+-45.11)/2=7.295 km/h East

Approximately 10 km/h East since it is positive

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y