12. A cyclist A started his journey on cycle at
7.30 a.m. at a speed 8 km/hr. B, another
cyclist, started from the same point half an
hour later but with a speed of 10 km/hr.
At what time did B overtake A ?
Answers
Answer:
Let B meets A after time t hr from the common starting point .
So, same distance is covered by A & B but B takes 30 min or o.5 hr less for his higher speed of 10 kmph over speed of A as 8 kmph .
The distance travelled by A in t hr is 8t km and B travelled in (t-0.5) hr is 10t km - 5 km
As both distance is equal 8t = 10t - 5, or
2t = 5 , or t = 2.5 hr or 2 hr 30 min so at time 10 AM+ , B crosses A
Step-by-step explanation:
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Given : A cyclist A started his journey on cycle at 7.30 a.m. at a speed 8 km/hr.
B, another cyclist, started from the same point half an hour later but with a speed of 10 km/hr.
To Find : At what time did B overtake A
Solution:
Method 1 :
A started his journey on cycle at 7.30 a.m. at a speed 8 km/hr.
Distance covered in 1/2 hr = (1/2) * 8 = 4 km
B started his journey at 8 AM
relative speed of B = 10 - 8 = 2 km/hr
relative distance to be covered = 4 km
Time taken = 4/2 = 2 hrs
8 + 2 = 10 AM
after 10 AM B overtake A
Method 2 ;
Let say after t hours of 8 AM when B started they meet
A traveled for t + 1/2 hr
B travelled for t hrs
Distance covered are equal
10t = 8(t + 1/2)
=> 2t = 4
=> t = 2
Hence 8 + 2 = 10 AM
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