Question

A ship moving with a constant acceleration of 25km/h2 in a fixed direction speeds up from 10km/h to
15km/h. Find the distance travel by the ship in this period.

Answer 1

u = 10kmph
v = 15kmph
a = 25kmph²

Just apply the 3rd equation of motion,

v²= u² + 2as
15² = 10² + 2× 25 × s
15²-10² = 50s
5×25 = 50s
s = 5×25/50 = 2.5 km

I hope you understand the approach.
All the best!

Answer 2

As we know that,
2as= v2- u2

As it is given in the question,
v= 15 km/h
u=10km/h
a= 25km/h2

by substituting the values we get,

2 as = v2 - u2
2 × 25 × s = ( 15 )( 15 ) -  ( 10 ) ( 10 )
50 × s =  225 -100
s= 125 /  50
:. Therefore , 
 s= 2.5km is the Answer.

Hope I helped in time.
Thank q