Question #da5cc

Oct 27, 2016

$0.2 s$, rounded to one decimal place.

Explanation:

Applicable kinematic equations are
${v}^{2} - {u}^{2} = 2 a s$ .....(1)
$v = u + a t$ .....(2)
where $s$ is distance traveled in time $t$, $u , v \mathmr{and} a$ are initial velocity, final velocity and acceleration of the object respectively.

Inserting given values in (1), we get (remember initial velocity of the dart is zero.)
${14}^{2} - {0}^{2} = 2 a \times 1.2$
$\implies a = {14}^{2} / \left(2 \times 1.2\right) = \frac{490}{6} m {s}^{-} 2$

Inserting calculated value of acceleration in (2) we get
$14 = 0 + \frac{490}{6} t$
$\implies t = \frac{14}{\frac{490}{6}}$
$\implies t = 0.2 s$, rounded to one decimal place.