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 and position 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 ac- celerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?

Key Concepts

Kinematic Equations

Kinematics involves the equations that relate displacement, velocity, and acceleration of an object. In problems like this, knowing one kinematic function allows you to derive others through differentiation or integration, which is fundamental in understanding motion in one dimension.

Velocity and its Derivatives

The velocity function describes how the speed of an object changes over time. Differentiating the velocity function with respect to time gives the acceleration, which quantifies the rate of change of velocity, while integrating the velocity function gives the position, or displacement, of the object over time.

Differentiation to Find Acceleration

Acceleration is exactly the derivative of the velocity function. Calculus is employed to differentiate the given velocity expression with respect to time, yielding a function that tells us how quickly the bullet's speed is changing as it travels down the barrel.

Integration for Position Calculation

Since position is the integral of the velocity function, applying integration (and using appropriate initial conditions) will determine the bullet’s displacement or the distance traveled while it is being accelerated within the barrel.

Identifying Extremum Conditions

In this context, setting the acceleration function equal to zero allows you to find the time at which the velocity is at an extremum (in this case, maximum). This is critical in determining when the bullet is no longer accelerating and is ready to exit the barrel.

Transcript

00:01 Hi everyone here it is given the speed of a bullet as it travels down the wearer of a rifle towards the opening is given by minus 5 into 10 to the power 7 t square plus 3 into 10 to the power 5 into t here speed is measured in meter per second and time is measured in second.

00:44 It is given that when it escapes the variant, acceleration is zero.

01:03 We have to calculate acceleration as a function of time, position as a function of time.

01:22 In b part, we have to find the time when it is escaping the the variant c the speed at which speed at which it leaves the warrant and d length of the variant we have to calculate first part acceleration is defined as rate of change of velocity with time so velocity as a function of time already we give be no substituting here minus 5 10 to the power 7 t squared plus 3 into 10 to the power 5 into t so differentiating this equation we will get the acceleration so acceleration of the bullet is minus 10 into 10 to the power 7 t plus 3 into 10 to the power 5 so this is the answer for acceleration as a function of time now if at t skull to 0 initial position of the bullet is 0 then we can find velocity is dx upon dt so dx you may write v into dt now integrating it under the condition 0 to x and 0 to t velocity you know substitute the value here velocity is given minus 5 into 10 to the power 7 t squared plus 3 into 10 to the power 5 into t it has to be integrated now integrating left and right hand side and so we will get x is equal to minus 5 by 3 into 10 to the power 7 t cube plus 3 by 2 into 10 to the power 5 t squared so x can be written as minus 1 .67 into 10 to the power 7 t cube plus 1 .5 into 10 to the power 5 t squared so this is the answer for part a in part b we have to find the time when it is leaving the barrel so at the time of bullet escaping the barrel acceleration is 0 so the bullet escape at zero acceleration you know the equation acceleration put the value you will get 0 is equal to minus 10 into 10 to the power 7 t plus 3 into 10 to the power 5 is called to 0 so t you will get 3 into 10 to the power minus 3 seconds so after this time bullet will leave the barrel see part ballacity at the time of leaving the barrel you can find substitute the value in the expression of velocity so you will get minus 5 into 10 to the power of 7 into t means 3 into 10 to the power minus 3 whole square plus you have 3 into 10 to the power minus 3 into t that is 3 into 10 to the power my 3 so this is plus 5 sorry 3 into 10 to 4 let me correct it so you will get the velocity to be 450 meter per second.

06:06 So, bullet leaving the barrel of the gun with the speed of 450 meter per second.

06:12 To calculate the length of the barrel, we have to find the distance covered before leaving the barrel...

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