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 length of time 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

Quadratic Equations

Quadratic equations often arise in motion problems when dealing with functions that describe acceleration, velocity, or position in a parabolic form. Solving these equations is essential for finding important time instants or other characteristics of the motion, such as when acceleration reaches a particular value or when an object reaches a specific position.

Integration

Integration is the inverse process of differentiation, used to accumulate quantities such as displacement over a period of time from a given velocity function. In kinematic problems, integrating the velocity function yields the position function, which represents the total displacement over time.

Differentiation

Differentiation is the mathematical process used to compute the rate of change of a function. In the context of kinematics, it allows one to determine the instantaneous velocity as the derivative of the position function, and the acceleration as the derivative of the velocity function, thereby revealing how motion evolves over time.

Calculus-Based Kinematics

This concept involves using calculus to analyze motion, establishing the mathematical relationships between displacement, velocity, and acceleration. It typically uses derivatives to go from position to velocity and from velocity to acceleration, and integration to determine displacement from velocity, providing a framework for understanding motion in a continuous way.

Transcript

00:01 In the part a, we have to find the acceleration.

00:04 We know that acceleration is the derivative of velocity function with respect to time.

00:10 So we have given the velocity is minus 5 .00 into 10 to the power 7, t squared plus 3 .00 into 10 to the power 5 into t.

00:27 After differentiating it, the value of acceleration as a function of t will become minus 10 .00 into 10 to the power 7 meter per second cube into t plus 3 .00 into 10 to the power 5 meter per second square.

00:55 Now we have to find the position function.

01:00 For the position function we know that v is equal to d x by d t.

01:04 From here we know that x initial is 0 at t equal to 0 then we will use d x is equal to v into d t again integrating it x minus 0 is equal to 0 to t we will put the value of velocity function here and after integrating it we will get the value of position function as a function of time t is minus 1 .67 into 10 to the power 7 meter per second cube into pq plus 1 .50 10 to the power 5 meter per second square into t square this is the position function now for part b we have to find we have to find the time for the time we will know that a acceleration value is zero initially minus 10 .00 10 to the power 7 we will equate the acceleration equal to 0 this is the value of acceleration as a function equal to 0 from here we will get the time the time will be equal to 3 .00 into 10 to the power 5 divided by 10 point 10 to the power 7 it will equal to 3 .00 into 10 to the power minus 3 .3 second.

02:56 This is the solution of part v.

03:00 For part c, we have to find the velocity.

03:04 So we will put the value of time in the velocity function...

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