A man walks backward on a bus. To a stationary observer, the bus moves
east at a speed of 15 m/s. If the man moves 2.3 m/s west relative to the bus,
what is his velocity in the frame of reference of a stationary observer?

Explanation:

V(m,b)= V(m,e) - V(b,e)

2.3m/s (west) = V(m,e) - 15m/s (east)

to solve this change the direction of slower speed. so it will become negative value

-2.3m/s ( east) = V (m,e) - 15m/s (east)

V(m,e) = 15m/s (east) - 2.3m/s (east)

V(m,e) = 12.7 m/s (east)

velocity of the man in frame of reference of stationary observer is 12.7 m/s

it's slightly less than the velocity of bus so stationary observer see the man move slower than the one inside the bus

ps:- V(m,b) - velocity of man relative to bus

V(b,e) - velocity of bus relative to earth

V(m,e) - velocity of man relative to earth

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2