Answer to Question #169141 in Physics for LYKA

Question

3. The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by

𝑣 = (−5.00 𝑥 10⁷) t² + (3.00 x 10⁵) t where v is in meters per second and t is in seconds. The

acceleration of the bullet just as it leaves the barrel is zero.

(a) Determine the acceleration

and position of the bullet as functions of time when the bullet is in the barrel.

(b) Determine the

time interval over which the bullet is accelerated.

(c) Find the speed at which the bullet leaves

the barrel.

(d) What is the length of the barrel?

Answer 1

a) In order to find acceleration of the bullet we need to take the derivative of $v$ with respect to $t$:

$$a=\dfrac{dv}{dt},$$ $$a=\dfrac{d}{dt}[(-5.0\cdot10^7)t^2+(3.0\cdot10^5)t]=-(10.0\cdot10^7\ \dfrac{m}{s^3})t+3.0\cdot10^5\ \dfrac{m}{s^2}.$$

By the definition, velocity is the derivative of position with respect to time:

$$v=\dfrac{dx}{dt}.$$

Then, we can find the position of the bullet by integrating $v$ over $t$:

$$x=\displaystyle\intop_{0}^t vdt,$$ $$x=\displaystyle\intop_{0}^t (-5.0\cdot10^7t^2+3.0\cdot10^5t)dt=-5.0\cdot10^7\dfrac{t^3}{3}+3.0\cdot10^5\dfrac{t^2}{2},$$ $$x=-(1.67\cdot10^7\ \dfrac{m}{s^3})t^3+(1.5\cdot10^5\ \dfrac{m}{s^2})t^2.$$

(b) The bullet leaves the barrel of the rifle when its acceleration equals zero:

$$a=0,$$ $$-(10.0\cdot10^7\ \dfrac{m}{s^3})t+3.0\cdot10^5\ \dfrac{m}{s^2}=0,$$ $$t=\dfrac{3.0\cdot10^5\ \dfrac{m}{s^2}}{10.0\cdot10^7\ \dfrac{m}{s^3}}=3\cdot10^{-3}\ s.$$

(c) As we know the time interval over which the bullet is accelerated, we can find the speed at which the bullet leaves the barrel:

$$v=-5.0\cdot10^7\ \dfrac{m}{s^3}\cdot(3\cdot10^{-3}\ s)^2+3.0\cdot10^5\ \dfrac{m}{s^2}\cdot3\cdot10^{-3}\ s=450\ \dfrac{m}{s}.$$

(d) Substituting time into the position equation of the bullet motion, we can find the length of the barrel:

$$x=-(1.67\cdot10^7\ \dfrac{m}{s^3})\cdot(3\cdot10^{-3}\ s)^3+(1.5\cdot10^5\ \dfrac{m}{s^2})\cdot(3\cdot10^{-3}\ s)^2=0.9\ m.$$