solve my question 17 plz
Attachments:
Answers
Answered by
0
Given
velocity of cycle = 5km/h or 5×1000/60
= 500/6 m/min
time = 10
therefore distance covered in first case = 500/6×10
= 5000/6 m
now as we know that V^2 = U^2+ 2as
and given - a = 1 km/h or 1×1000/60
= 100/6
therefore
0 = (500/6)^2 - 2×100/6 × s
-2500/36 = -200/6 ×s
(2500/36)/(200/6) = s
2500/36× 6/200 = s
=25/12
THEREFORE total distance = 5000/6 + 25/12
= 10025/12
velocity of cycle = 5km/h or 5×1000/60
= 500/6 m/min
time = 10
therefore distance covered in first case = 500/6×10
= 5000/6 m
now as we know that V^2 = U^2+ 2as
and given - a = 1 km/h or 1×1000/60
= 100/6
therefore
0 = (500/6)^2 - 2×100/6 × s
-2500/36 = -200/6 ×s
(2500/36)/(200/6) = s
2500/36× 6/200 = s
=25/12
THEREFORE total distance = 5000/6 + 25/12
= 10025/12
tony0522:
wrong h bro ans
answer kya he
i got answer it is 40/3 km
Similar questions