A gives B a start of 30m and beats him by 5 sec in a race of 300m . On another day he gives start A gives a start of 50m to B and still beats him by 1 sec. Find the velocity of B.(in m/s) (Assume both have been travelling at the constant speed through out)
Answers
GIVEN
- A gives B a start of 30m and beats him by 5 sec in a race of 300m
- A gives a start of 50m to B and still beats him by 1 sec.
TO FIND
- Find the velocity of B. (in m/s)
Concepts Used
- Time, Speed & Distance.
SOLUTION
- Let the total time taken by B be x
- Speed = Distance / Time
ATQ
- Speed of A = 300 / x - 5
- Speed of B = 270 / x
- Speed of A = 300 / x - 1
- Speed of B = 250 / x
- [300 / x - 5] ÷ [270 / x] = [300 / x - 1] ÷ [250 / x]
- [300 / x - 5] × [x / 270] = [300 / x - 1] × [x / 250]
- 10x / 9x - 45 = 6x / 5x - 5
- 50x² - 50x = 54x² - 270x
- 4x² = 220x
- 4x = 220
- x = 55
- Velocity of B = 4.9 ~ 4.5 m/s
Solution :-
Let time taken by b to complete the race is x seconds.
Case 1) :- A gives B a start of 30m and beats him by 5 sec in a race of 300m .
→ Distance covered by A = 300 m
→ Time taken by A = (x - 5) seconds .
so,
→ Speed of A = D/T = 300/(x - 5) m/s
and,
→ Distance covered by B = 300 - 30 = 270 m
→ Time taken by B = x seconds .
so,
→ Speed of B = D/T = (270/x) m/s
Case 2) :- A gives a start of 50 m to B and still beats him by 1 sec.
→ Distance covered by A = 300 m
→ Speed of A = 300/(x - 5) seconds .
→ Time taken by A = D/S = 300 * (x - 5)/300 = (x - 5) seconds .
and,
→ Distance covered by B = 300 - 50 = 250 m
→ Speed of B = (270/x) m/s
→ Time taken by B = D/S = 250 * (x/270) = (25x/27) seconds .
A/q,
→ (25x/27) - (x - 5) = 1
→ (25x - 27x + 135) / 27 = 1
→ 135 - 2x = 27
→ 2x = 108
→ x = 54 seconds .
therefore,
→ Speed of B = 270/54 = 5 m/s (Ans.)
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