A cyclist starts from the centre O of a circular track of radius 1 km, reaches the edge A of the track and then cycles along the circumference and stops at point B. If the total time taken is 10 minutes, what is the a) average velocity b) average speed of the cyclist. 

a) Displacement in 10 min (=1/6 hr) = OB = r = 1 km

Therefore, 

Average velocity = 
b) Distance travelled in 1/6 hr = 0 to A (along radius)  + A to B (along the arc)

                                              = r + 

                                              = 1 km + 1 km + 

                                               = 2.05 km

Therefore, 

Averge speed =  = 12.30 km/hr

Relative velocity of one train w.r.t second, 

= 42 - (-30)

= 72 km /hr

= 20 m/s 

Total distance to be travelled = 120 + 80 = 200 m

Time taken  = 200/20 = 10 sec

Here, distance between two towns A and B, S = 480 km 

v₁ = S/8   and
 
v₂ = S/10

If these buses meet after time t,  then 

S/8 x t + S/10 x t = S 

On solving, we get 

t = 40/9 hours 

Therefore, distance from A = 480/80 x 40/9 = 800/3 = 266.7 km 

Total distance to be travelled  for overtaking the car = 5.0  +5.0 = 10. 0 m 

Relative velocity of first car w.r.t second car = 60 - 42 = 8 km/hr

     = 8 x 1000/ (60 x 60)

      = 50 m/s

Time taken, t = 10/5 = 2 s.

Road distance used for overtaking = distance travelled by first car in 2 s + length of first car

= [ 60 x 1000 / (60 x60)] x 2+ 5

= 38.33 m