A ship moving with a constant acceleration of 36km/h2 in a fixed direction speeds up from 18km/h to 36km/h. Find the distance traversed by the ship in this period.
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EXPLANATION.
- GIVEN
Ship moving with a constant acceleration
of 36 km/hr² .
speed up from 18km/hr to 36km/hr.
To find the distance traverse by the
ship in this period.
according to the question,
Acceleration = 36 km/hr²
initial velocity = u = 18 km/hr
final velocity = v = 36 km/hr
From Newton third equation of
kinematics.
=> v² = u² + 2as
=> ( 36)² = ( 18)² + 2 ( 36 ) s
=> ( 36² - 18² ) = 72s
=> ( 36 + 18 ) ( 36 - 18 ) = 72s
=> 54 X 18 = 72s
=> s = 54/4
=> s = 13.5 km
Therefore,
Distance traverse by the ship in this
period = 13.5 km
Related formula.
1) = Newton first equation of
kinematics.
=> v = u + at
2) = Newton second equation of
kinematics.
=> s = ut + 1/2 at²
3) = Newton third equation of
kinematics.
=> v² = u² + 2as
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