Question

A ship moving with a constant acceleration of $36km/h^2$ in a fixed direction speeds up from $12km/h$ to $18km/h.$ Find the distance traversed by the ship in this period.

Answer 1

Step 1: Given data:

The constant acceleration in the ship($a$) = $36km{h}^{-2}$

Initial velocity($u$) of the ship = $12km{h}^{-1}$

Final velocity of the ship($v$) = $18km{h}^{-1}$

Step 2: Calculation of the distance covered by the ship:

Using the third equation of motion; $v^2=u^2+2as$

Where, $v$ = Final velocity

$u$ = Initial velocity

$s$ = Distance covered by the body

$a$ = Acceleration in the body

Let $s$ be the distance traversed by the ship.

(18)2−(12)2=2(36)s {{(18)}^{2}}-{{(12)}^{2}}=2(36)s Putting the values, we get

${\left(18km{h}^{-1}\right)}^2={\left(12km{h}^{-1}\right)}^2+2\times 36km{h}^{-2}\times s$

$324=144+72\times s$

$\Rightarrow 72\times s=324-144$

$\Rightarrow 72\times s=180$

$\Rightarrow s=\frac{180}{72}$

$\Rightarrow s=2.5km$

Thus,

The distance traversed by the ship is $2.5km$ .