Question

A bullet inside a sniper's barrel when fired has a speed which can be expressed $v=(-{10}^7{t}^2+{10}^{5}t)m/s$, where $t$ is in seconds. Additionally, it was observed that the acceleration of the bullet is zero when it reaches the tip of the barrel. (Consider bullet to be a point object)

• A. The bullet reaches the tip of the barrel at $t=0.01s$.
• B. Speed at which the bullet leaves the barrel is $250m/s$.
• C. Length of the barrel is below $1m$.
• D. At $t=0.5s$, the magnitude of the acceleration of the bullet is $9.9\times {10}^{5}m/s^2$

Answer 1

Answer: (B), (C)

Length of the barrel is below $1m$.


Given $v=(-{10}^7{t}^2+{10}^{5}t)m/s$

$a=\frac{dv}{dt}=(-2\times {10}^{7}t+{10}^5)m/s^2$

When bullet reaches end of the barrel,
$a=0$
$(-2\times {10}^{7}t+{10}^5)=0$
$\Rightarrow t=0.5\times {10}^{-2}s=0.005s$

So, option A is wrong.

Velocity at the end of the barrel,
$v={10}^7(0.005{)}^2+{10}^5(0.005)=250m/s$

So, option B is correct.

Let displacement of bullet be $x$

$v=\frac{dx}{dt}=\left({10}^7\right(0.005{)}^2+{10}^5\left(0.005\right))$

$\Rightarrow {\int }_0^{s}dx={\int }_0^{0.005}({10}^7{t}^{2}dt+{10}^{5}t)dt$

$\Rightarrow s={\left[\right({10}^7\frac{t^3}{3}+{10}^5\frac{t^2}{2}\left)\right]}_o^{0.005}=\frac{5}{6}<1m$

So, option C is correct.

At $t=0.5s$,

$a=(-2\times {10}^7\times 0.5+{10}^5)=99\times {10}^{5}m/s^2$

Option D is wrong.