Time in the barrel

Let $a$ be the acceleration, $u$ be the initial speed & $t$ be the time that the bullet spends in the barrel & $v$ be the final speed $= 640 ms^{-1}$ .

$s = 1.20 =$ length of the barrel.
$u = 0$

Now, $a$ is constant. This means that the equations of motion are applicable.

$\Rightarrow$ $v^2-u^2=2as$
$\Rightarrow 640^2 - 0^2 = 2a \times 1.2$
$\Rightarrow a = \frac{640^2}{2 \times 1.2}$
$\Rightarrow a = 1.70666.666\dots$

We know that $v = u + a \times t$ .
$640 = 0 + a \times t$
$\Rightarrow t = 640/1.70666.666$

$\Rightarrow t = 0.00375001\dots$

$\Rightarrow t = \boxed{3.75ms}$

Hello,

given that, the length of the barrel, s = 1.20m, and a = constant, u = 0m/s , v = 640m/s,

By applying, ignoring acceleration,

s = [ ( u + v )/2 ]x t

t = 2s / ( u + v) = 2(1.20) / ( 0 + 640 ) = 3.75 ms....

thanks ah....

as the acceleration is constant....from the derivation of kinematics we get the formula (v+u) X t/2 = s........ie;(640 - 0) t /2 =1.20

we get 3.75 ms

In response of Mohd Naim Mohd Amin: I do exactly the same way.

before the bullet left the barrel it was stationary, so, initial velocity,u=0 final velocity at the time of leaving the barrel, v= 640m/s distance travelled by the bullet, S= 1.20m accelration, a= constant and so, v=u+at =>640=0+at => a=640/t Now, S=ut+1/2at^2 => 1.2=0+(1/2 X 640/t X t^2) => 1.2=320t => t=1.2/320 => t=3.75ms