A ship moving with a constant acceleration of 25km/h2 in a fixed direction speeds up from 10km/h to15km/h. Find the distance travel by the ship in this period.
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Acceleration (a)= 25km/h^2
Initial velocity (u)= 10km/h
Final velocity (v) =15km/h
Let distance travelled be s
Using third equation of motion
2as =v^2 -u^2
2 (25)(s) = (15)^2 - (10)^2
50s = 225 -100
50s = 125
s= 125/50
= 2.5
So the distance covered by that body is 2.5 km.
Initial velocity (u)= 10km/h
Final velocity (v) =15km/h
Let distance travelled be s
Using third equation of motion
2as =v^2 -u^2
2 (25)(s) = (15)^2 - (10)^2
50s = 225 -100
50s = 125
s= 125/50
= 2.5
So the distance covered by that body is 2.5 km.
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