The speed of a bullet as it travels down the barrel of a...

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by $v=\left(-5.00 \times 10^{7}\right) t^{2}+$ $\left(3.00 \times 10^{5}\right) t,$ 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 a function 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?

Key Concepts

Critical Points and Physical Interpretation

In the context of motion analysis, critical points occur when the derivative of a function (e.g., acceleration as the derivative of velocity) is zero. Recognizing these points enables one to identify key moments in the object's trajectory, such as when acceleration ceases, signaling changes in the motion behavior. This understanding is essential for determining conditions like the exit speed from a barrel or the extent of acceleration.

Quadratic Functions and Their Properties

Quadratic functions often appear in problems involving accelerated motion. Their properties, such as parabolic shape and easily findable extrema (maximum or minimum points), make them useful for modeling situations where acceleration varies with time. Understanding these properties helps in determining critical time points and values, such as the point where the acceleration becomes zero.

Kinematic Relationships in One Dimension

Kinematics in one dimension concerns the relationship between displacement, velocity, and acceleration. This subject uses calculus tools to connect these quantities, providing methods to calculate one when another is known. These relationships are crucial for analyzing motion scenarios where the speeds and accelerations are given as functions of time.

Differentiation in Kinematics

This concept involves taking the derivative of a given function to obtain another related physical quantity. In kinematics, differentiating the velocity function with respect to time gives the acceleration, which describes how the velocity changes over time. This relationship is fundamental for understanding the dynamics of objects in motion.

Integration in Kinematics

Integration is used to determine the position of an object when its velocity as a function of time is known. By integrating the velocity function with respect to time, one obtains the displacement or position function, which includes a constant of integration representing initial conditions. This process connects the rate of motion to the total distance or displacement covered.

Transcript

00:01 So we have an expression for the speed of the bullet when it exits at the barrel in terms of time.

00:08 To find firstly the acceleration a, we know that we should take the time derivative of this expression.

00:16 So a is equal to dv d t and this is d d d t of the expression above, minus 5 times 10 to the 7 t squared plus 3 times 10 to the 5 times t times t and if we take the time derivative we get the acceleration to be minus 10 times 10 to the 7 meters per second cubed times t plus 3 times 10 to the 5 meters per square second and hence we have an expression for acceleration.

01:20 Now, for our displacement, we know that let's take the initial position, xi, to be zero, at time t is equal to zero.

01:35 And we know that speed v is equal to the rate of change or displacement, dx, dt.

01:44 And so if we integrate both sides we get that x minus 0 is equal to the integral over time from zero to t of z d t and we can expand this integral that's the integral from zero to t for the expression that we know b md 5 times 10 to 7 t squared plus 3 times 10 to the 5 multiplied by t all that with respect to d t and if we perform this integration we get x to be minus 1 .67 times 10 to the 7 multiplied by t cubed plus 1 .5 times 10 to the 7 multiplied by t cubed plus 1 .5 times 10 to the 5 multiplied by t cubed plus 1 .5 times 10 to the 5 multiplied by t squared and these are all their si units so hence we have an expression for the position in terms of time as well as the acceleration in terms of time next we want to calculate time that the bullet spins in the barrel the time that the bullet spins in the barrel is the time that the bullet is being accelerated so the bullet escapes when a is equal to zero there's no acceleration on the bullet and this happens when our expression for acceleration is equal to 0...

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