A ship moving with a constant acceleration of 36 km/h² in a fixed direction speed up from 12 km/h to 18 km/h . Find the distance traversed by the ship in this period
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Answered by
17
2.5 km distance covered
v =u+at
18=12+36t,t=1/6 hour or 10 minute
s=ut+1/2 at square
s=2.5 km answer
v =u+at
18=12+36t,t=1/6 hour or 10 minute
s=ut+1/2 at square
s=2.5 km answer
sumitkumar8:
But this type of formula is not any where
Answered by
7
A = 10 m/s2
Vi = 3.33m/s
Vf= 5m/s
2 as = vf2- vi2
2 × 10 × s = (5)2. - (3.33)2
20× s = 25- 11.11
20× S = 13.89
S = 13.89 ÷ 20
S= 0.6944 m
Or 694.4 km
Vi = 3.33m/s
Vf= 5m/s
2 as = vf2- vi2
2 × 10 × s = (5)2. - (3.33)2
20× s = 25- 11.11
20× S = 13.89
S = 13.89 ÷ 20
S= 0.6944 m
Or 694.4 km
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