Question

a ship moving with a constant acceleration of 25km/h^2 in a fixed direction speeds up from 10km /h to 15 km/h . Find the distance travelled by a ship in this period

Answer 1

we know that,
2as= v2- u2

here,
v= 15 km/h
u=10km/h
a= 25km/h2

by substituting the values we get,

2as = v2 - u2
2 × 25× s= (15)(15)- (10)(10)
50 × s =  225-100
s= 125 / 50
therefore , 
s= 2.5 km

Answer 2

"The distance travelled by the ship is 2.5km.

Given:

Acceleration of the ship, $a = 25 \frac{km}{h^{2}}$

Initial velocity of the ship, $u = 10 \frac{km}{h}$

Final velocity of the ship,$v = 15 \frac{km}{h}$

Distance travelled by the ship, s = ?

Solution:

We know that the distance travelled by an object subject to constant acceleration is given by

$s=\frac{v^{2}-u^{2}}{2 a}$

Substitution

$s=\frac{15^{2}-10^{2}}{2 \times 25}$

$s=\frac{225-100}{50}$

$s=2.5 \mathrm{km}$

Acceleration:

Acceleration is aprocess of moving faster or happening more quickly."