Question

a revolving rifle fires 9 - mm bullets such that as they travel down the barrel of the rifle their speed is given by v = (-5.00×10⁷)t²+(2.50×10⁵)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 (in m/s²) and position (in m) of the bullet as a function of time when the bullet is in the barrel (use the following as necessary: t. round all numerical coefficients to at least three significant figures. assume that the position of the bullet at t = 0 is zero.) a(t) = m/s² x(t) = m

Explanation:

Step1: Recall the relationship between acceleration and velocity

Acceleration $a(t)$ is the derivative of velocity $v(t)$ with respect to time $t$, i.e., $a(t)=\frac{dv(t)}{dt}$. Given $v(t)=(- 5.00\times10^{7})t^{2}+(2.50\times10^{5})t$.

Step2: Differentiate $v(t)$ using the power - rule

The power - rule for differentiation is $\frac{d}{dt}(x^{n})=nx^{n - 1}$.
\[

$$\begin{align*} a(t)&=\frac{d}{dt}[(-5.00\times 10^{7})t^{2}+(2.50\times 10^{5})t]\\ &=(-5.00\times 10^{7})\times2t+(2.50\times 10^{5})\times1\\ &=(-1.00\times 10^{8})t + 2.50\times 10^{5} \end{align*}$$

\]

Step3: Recall the relationship between position and velocity

Position $x(t)$ is the integral of velocity $v(t)$ with respect to time $t$, i.e., $x(t)=\int v(t)dt$.
\[

$$\begin{align*} x(t)&=\int[(-5.00\times 10^{7})t^{2}+(2.50\times 10^{5})t]dt\\ &=(-5.00\times 10^{7})\frac{t^{3}}{3}+(2.50\times 10^{5})\frac{t^{2}}{2}+C \end{align*}$$

\]
Since $x(0) = 0$, then $C = 0$. So $x(t)=(-1.67\times 10^{7})t^{3}+(1.25\times 10^{5})t^{2}$

Answer:

$x(t)=(-1.67\times 10^{7})t^{3}+(1.25\times 10^{5})t^{2}\text{ m}$, $a(t)=(-1.00\times 10^{8})t + 2.50\times 10^{5}\text{ m/s}^2$