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 \(x =\) (9.0 \(\times\) 10\(^3\) m/s\(^2)t^2\) - (8.0 \(\times\) 10\(^4\) m/s\(^3)t^3\). 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

The gun barrel is 4.375 m long, speed is 300 m/s, forces are 2.7 × 10⁴ N at t = 0 and 9.0 × 10³ N at t = 0.025 s.

Step-by-step solution

Understanding the Displacement Function

The displacement function of the chicken-sized object is given as \(x(t) = (9.0 \times 10^3 \text{ m/s}^2)t^2 - (8.0 \times 10^4 \text{ m/s}^3)t^3\). This function describes how the position along the gun barrel changes with respect to time \(t\).

Calculating the Length of the Gun Barrel

The length of the gun barrel will be the displacement \(x\) at the time \(t = 0.025\) s. Substitute \(t = 0.025\) s into the displacement equation: \[ x = (9.0 \times 10^3)(0.025)^2 - (8.0 \times 10^4)(0.025)^3 \] Calculating gives: \[ x = (9.0 \times 10^3)(0.000625) - (8.0 \times 10^4)(0.000015625) \] \[ x = 5.625 - 1.25 = 4.375 \text{ m} \] So, the gun barrel must be 4.375 meters long.

Finding the Speed of the Objects

To find the speed, differentiate the displacement function \(x(t)\) with respect to time \(t\) to get the velocity function \(v(t)\): \[ v(t) = \frac{d}{dt}[(9.0 \times 10^3)t^2 - (8.0 \times 10^4)t^3] = 2(9.0 \times 10^3)t - 3(8.0 \times 10^4)t^2 \] \[ v(t) = (1.8 \times 10^4)t - (2.4 \times 10^5)t^2 \] Substitute \(t = 0.025\) s into the velocity equation: \[ v(0.025) = (1.8 \times 10^4)(0.025) - (2.4 \times 10^5)(0.025)^2 \] \[ v(0.025) = 450 - 150 = 300 \text{ m/s} \] Thus, the speed of the objects as they leave the gun is 300 m/s.

Calculating the Net Force at t = 0

Find acceleration by differentiating the velocity function \(v(t)\): \[ a(t) = \frac{d}{dt}[(1.8 \times 10^4)t - (2.4 \times 10^5)t^2] = 1.8 \times 10^4 - 4.8 \times 10^5 t \] At \(t = 0\), substitute: \[ a(0) = 1.8 \times 10^4 \text{ m/s}^2 \] Use Newton's second law to calculate force: \[ F = ma = (1.50 \text{ kg})(1.8 \times 10^4) = 2.7 \times 10^4 \text{ N} \] Thus, the net force at \(t = 0\) is \(2.7 \times 10^4 \text{ N}\).

Calculating the Net Force at t = 0.025 s

Using the acceleration formula \(a(t) = 1.8 \times 10^4 - 4.8 \times 10^5 t\), substitute \(t = 0.025\): \[ a(0.025) = 1.8 \times 10^4 - 4.8 \times 10^5 (0.025) \] \[ a(0.025) = 1.8 \times 10^4 - 1.2 \times 10^4 = 6.0 \times 10^3 \text{ m/s}^2 \] Use Newton's second law again: \[ F = ma = (1.50)(6.0 \times 10^3) = 9.0 \times 10^3 \text{ N} \] Therefore, the net force at \(t = 0.025\) s is \(9.0 \times 10^3 \text{ N}\).

Key concepts

Kinematic Equations

Kinematics deals with the motion of objects and helps us understand how they move through space. The kinematic equations are mathematical formulas used primarily to calculate an object's displacement, velocity, and acceleration over time.
In the original problem, the displacement function is given as:

• \( x(t) = (9.0 \times 10^3 \text{ m/s}^2)t^2 - (8.0 \times 10^4 \text{ m/s}^3)t^3 \).

This function provides valuable information about how the object's position changes over time while it moves through the gun barrel. By integrating the basic principles of calculus, you can derive velocity and acceleration functions from the displacement function.
To find the velocity, you differentiate the displacement function with respect to time, obtaining:

• \( v(t) = \frac{d}{dt}[(9.0 \times 10^3)t^2 - (8.0 \times 10^4)t^3] = (1.8 \times 10^4)t - (2.4 \times 10^5)t^2 \).

For acceleration, further differentiation of the velocity function gives:

• \( a(t) = \frac{d}{dt}[(1.8 \times 10^4)t - (2.4 \times 10^5)t^2] = 1.8 \times 10^4 - 4.8 \times 10^5 t \).

These kinematic equations help calculate crucial factors, like the speed of the object at a specific point in time in the barrel, and are essential for solving physics problems involving motion.

Newton's Second Law

Newton's second law of motion is a powerful tool in physics that relates the net force acting on an object to its mass and acceleration. The law is expressed as:

• \( F = ma \),

where:

• \( F \) is the net force,
• \( m \) is the object's mass, and
• \( a \) is the acceleration.

In the exercise, applying Newton's second law reveals how much force is needed to accelerate the 1.50-kg object in the gun barrel. First, at \( t = 0 \), the acceleration is calculated as \( 1.8 \times 10^4 \text{ m/s}^2 \), leading to a net force:

• \( F = (1.50 \text{ kg})(1.8 \times 10^4 \text{ m/s}^2) = 2.7 \times 10^4 \text{ N} \).

At \( t = 0.025 \) s, the acceleration changes to \( 6.0 \times 10^3 \text{ m/s}^2 \), resulting in a net force of:

• \( F = (1.50 \text{ kg})(6.0 \times 10^3 \text{ m/s}^2) = 9.0 \times 10^3 \text{ N} \).

Understanding these changes in force is crucial for engineering applications, like designing devices capable of similar tasks, such as the test gun in question.

Displacement and Velocity

In kinematics, displacement refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. Velocity, on the other hand, describes how fast an object is moving and the direction of movement; it too is a vector quantity.
In the context of this problem, the displacement equation \( x(t) = (9.0 \times 10^3 \text{ m/s}^2)t^2 - (8.0 \times 10^4 \text{ m/s}^3)t^3 \) describes how the object's position changes as time progresses inside the gun barrel.
To find the velocity, the derivative of the displacement function is taken. This yields:

• \( v(t) = (1.8 \times 10^4)t - (2.4 \times 10^5)t^2 \).

This equation gives the instantaneous velocity at any point in time by substituting values for \( t \). When the object reaches the end of the barrel at \( t = 0.025 \) s, the velocity is calculated to be 300 m/s.
Understanding these concepts assists in anticipating the movement of objects, important for applications not only in testing but also in broader fields such as automotive and flight dynamics.