Question

To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by $\mathbf{x}\mathbf{=}\left(9.0\times {10}^{3}m/s^2\right){\mathbf{t}}^{\mathbf{2}}\mathbf{-}\left(8.0\times {10}^{4}m/s^2\right){\mathbf{t}}^{\mathbf{3}}$. The object leaves the end of the barrel at t=0.025 s. (a) How long must the gun barrel be? (b) What will be the speed of the objects as they leave the end of the barrel? (c) What net force must be exerted on a 1.50-kg object at (i) t=0 and (ii) t=0.025 s ?

Short answer

(a) The length of the gun barrel is 4.375 m.

(b)The speed of the objects is 300 m/s.

(c) The net force exerted at $t=0$is $2.7\times {10}^3\mathrm{N}$.

The net force exerted at t=0.025 s is$9\times {10}^3\mathrm{N}$ .

Step-by-step solution

Identification of the given data

The given data is listed below as:

• The time taken by the object to leave the barrel is,t=0.025 s
• The equation of the displacement is given as,$x=\left(9.0\times {10}^{3}m/s^2\right)t^2-\left(8.0\times {10}^{4}m/s^3\right)t^3$
• The mass of the object is, m=1.50 kg

Significance of the velocity and acceleration 

The velocity is described as the derivative of the distance with respect to time. The acceleration is described as the derivative of the velocity with time.

(a) Determination of the length of the gun barrel

The equation of the length of the gun barrel is expressed as:

$x=\left(9.0\times {10}^{3}m/s^2\right)t^2-\left(8.0\times {10}^{4}m/s^2\right)t^3$ …(i)

Here,tis thetime taken by the object to leave the barrel.

Substitute 0.025 s for t in the above equation.

$$\begin{gathered} \mathrm{x}=\left(9.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2\right){\left(0.025\mathrm{s}\right)}^2-\left(8.0\times {10}^4\mathrm{m}/{\mathrm{s}}^3\right){\left(0.025\mathrm{s}\right)}^3 \\ =\left(9.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2\right)\left(6.25\times {10}^{-4}{\mathrm{s}}^2\right)-\left(8.0\times {10}^4\mathrm{m}/{\mathrm{s}}^3\right)\left(1.5625\times {10}^{-5}{\mathrm{s}}^3\right) \\ =5.625\mathrm{m}-1.25\mathrm{m} \\ =4.375\mathrm{m} \end{gathered}$$

Thus, the length of the gun barrel is 4.375 m.

(b) Determination of the speed of the objects

The velocity of the objects is being gathered by differentiating the equation (i) with respect to the time t.

$v=\frac{dx}{dt}=\left(18.0\times {10}^{3}m/s^2\right)t-\left(24.0\times {10}^{4}m/s^3\right)t^2$ …(ii)

Substitute 0.025 s fort in the above equation.

$$\begin{gathered} \mathrm{v}=\left(18.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2\right)\left(0.025\mathrm{s}\right)-\left(24.0\times {10}^4\mathrm{m}/{\mathrm{s}}^3\right){\left(0.025\mathrm{s}\right)}^2 \\ =\left(450\mathrm{m}/\mathrm{s}\right)-\left(24.0\times {10}^4\mathrm{m}/{\mathrm{s}}^3\right)\left(6.25\times {10}^{-4}{\mathrm{s}}^2\right) \\ =450\mathrm{m}/\mathrm{s}-150\mathrm{m}/\mathrm{s} \\ =300\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the objects is 300 m/s.

(c) Determination of the net force exerted at t=0

The acceleration of the objects is being gathered by differentiating the equation (ii) with respect to the time t.

$a=\frac{dv}{dt}=\left(18.0\times {10}^{3}m/s^2\right)-\left(48.0\times {10}^{4}m/s^3\right)t$

Substitute 0 for t in the above equation.

$a=18.0\times {10}^{3}m/s^2$

The equation of the net force is expressed as:

F =ma

Here,m is the mass anda is the acceleration of the objects.

Substitute 1.50 kg form and$18.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2$ fora in the above equation.

$$\begin{gathered} \mathrm{F}=\left(1.50\mathrm{kg}\right)\left(18.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2\right) \\ =2.7\times {10}^3\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =2.7\times {10}^3\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =2.7\times {10}^3\mathrm{N} \end{gathered}$$

Thus, the net force exerted at t =0 is $2.7\times {10}^{3}N$.

(c) Determination of the net force exerted at t=0.025 s  

The acceleration of the objects is being gathered by differentiating the equation (ii) with respect to the time t.

$a=\frac{dv}{dt}=\left(18.0\times {10}^{3}m/s^2\right)-\left(48.0\times {10}^{4}m/s^3\right)t$

Substitute 0.025 s for t in the above equation.

$$\begin{gathered} \mathrm{a}=18.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2-\left(48.0\times {10}^4\mathrm{m}/{\mathrm{s}}^3\right)\left(0.025\mathrm{s}\right) \\ =18.0\times {10}^3\mathrm{m}/{\mathrm{s}}^2-12000\mathrm{m}/{\mathrm{s}}^2 \\ =6\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The equation of the net force is expressed as:

F = ma

Here, F is the net force, m is the mass and a is the acceleration of the objects.

Substitute 1.50-kg for m and $6\times {10}^3\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$$\begin{gathered} \mathrm{F}=\left(1.50\mathrm{kg}\right)\left(6\times {10}^3\mathrm{m}/{\mathrm{s}}^2\right) \\ =9\times {10}^3\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =9\times {10}^3\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =9\times {10}^3\mathrm{N} \end{gathered}$$

Thus, the net force exerted at T=0.025 s is $9\times {10}^3\mathrm{N}$.