Two buses started simultaneously towards each other from towns A and B which are 480 km apart. It took the first bus travelling from A to B eight hours to cover the distance and the second bus travelling from B to A, ten hours. Determine when the buses will meet after starting and at what distance from A.
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
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
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
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