a projectile enters a resisting medium at x = 0 with an initial velocity v0 = 900 ft/s and travels 4 in…

a projectile enters a resisting medium at x = 0 with an initial velocity v0 = 900 ft/s and travels 4 in. before coming to the rest. assuming that the velocity of the projectile is defined by the relation v = v0 - kx, where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9 in. into the resisting medium.

a projectile enters a resisting medium at x = 0 with an initial velocity v0 = 900 ft/s and travels 4 in. before coming to the rest. assuming that the velocity of the projectile is defined by the relation v = v0 - kx, where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9 in. into the resisting medium.

Answer

Explanation:

Step1: Find the value of k

When the projectile comes to rest, $v = 0$. The initial velocity $v_0=900$ ft/s and it travels $x = 4$ in $=\frac{4}{12}=\frac{1}{3}$ ft before coming to rest. Substitute into $v = v_0 - kx$: [0 = 900 - k\times\frac{1}{3}] [k\times\frac{1}{3}=900] [k = 2700]

Step2: Find the initial acceleration

Acceleration $a=\frac{dv}{dt}$. Since $v = v_0 - kx$, differentiating with respect to time $t$ using the chain - rule $\frac{dv}{dt}=\frac{dv}{dx}\times\frac{dx}{dt}=v\frac{dv}{dx}$. Differentiating $v = v_0 - kx$ with respect to $x$ gives $\frac{dv}{dx}=-k$. At $x = 0$, $v = v_0$. So the initial acceleration $a_0=v_0\times(-k)$. Substitute $v_0 = 900$ ft/s and $k = 2700$: [a_0=900\times(- 2700)=-2.43\times10^{6}\text{ ft/s}^2]

Step3: Find the velocity as a function of time

We know that $v=\frac{dx}{dt}=v_0 - kx$. Rearranging gives $\frac{dx}{v_0 - kx}=dt$. Integrating both sides: [\int_{0}^{x}\frac{dx}{v_0 - kx}=\int_{0}^{t}dt] Let $u = v_0 - kx$, then $du=-kdx$. When $x = 0$, $u = v_0$; when $x=x$, $u = v_0 - kx$. [-\frac{1}{k}\int_{v_0}^{v_0 - kx}\frac{du}{u}=t] [-\frac{1}{k}[\ln(u)]_{v_0}^{v_0 - kx}=t] [t=\frac{1}{k}\ln\left(\frac{v_0}{v_0 - kx}\right)]

Step4: Convert 3.9 in to feet

$x = 3.9$ in $=\frac{3.9}{12}=0.325$ ft.

Step5: Calculate the time required

Substitute $v_0 = 900$ ft/s, $k = 2700$ and $x = 0.325$ ft into $t=\frac{1}{k}\ln\left(\frac{v_0}{v_0 - kx}\right)$: [t=\frac{1}{2700}\ln\left(\frac{900}{900-2700\times0.325}\right)] [t=\frac{1}{2700}\ln\left(\frac{900}{900 - 877.5}\right)=\frac{1}{2700}\ln\left(\frac{900}{22.5}\right)=\frac{1}{2700}\ln(40)] [t=\frac{\ln(40)}{2700}\approx\frac{3.6889}{2700}\approx1.366\times10^{-3}\text{ s}]

Answer:

(a) The initial acceleration of the projectile is $-2.43\times 10^{6}\text{ ft/s}^2$. (b) The time required for the projectile to penetrate 3.9 in. into the resisting medium is approximately $1.366\times 10^{-3}\text{ s}$.