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Your dorm wishes to purchase a small used cannon with the intent of firing a blank after each touchdown at home games. The spec sheet for a cannon of interest is missing some pages; however, the information you have claims that when the cannon is fired, the position of the cannonball as a function of time (while it is in the barrel of the cannon) is given by:

x(t) = (1.70 ✕ 10^5 m/s^2)t^2 − (1.20 ✕ 10^7 m/s^3)t^3.

The spec sheet further claims that the cannon design is such that the acceleration of the cannonball is zero at the exit end of the barrel. In order to learn more about the cannon, you are asked to determine the following.

a. time it takes for a cannonball to travel the length of the barrel
How can you determine an expression for the acceleration of the cannonball from an expression for the position of the cannonball? What is the acceleration of the cannonball as it exits the barrel of the cannon? s
b. speed of a cannonball as it exits the barrel
How can you determine an expression for the velocity of the cannonball from an expression for the position of the cannonball? How are the speed and velocity of the cannonball related? At what time do you wish to determine the speed of the cannonball? m/s
c. length of the barrel

Respuesta :

skyluke89

a) [tex]t=4.7\cdot 10^{-3} s[/tex]

b) [tex]v=802.7 m/s[/tex]

c) L = 2.51 m

Explanation:

a)

The position of the cannonball at time t is given by

[tex]x(t)=(1.70\cdot 10^5)t^2 -(1.20\cdot 10^7)t^3[/tex]

This is the position of the cannonball inside the barrel, measured with respect to the starting point.

The velocity of the cannonball can be found by calculating the derivative of the position; we find:

[tex]v(t)=x'(t)=2(1.70\cdot 10^5)t-3(1.20\cdot 10^7)t^2=\\=3.40\cdot 10^5 t-3.60\cdot 10^7 t^2[/tex]

Similarly, the acceleration of the cannonball is given by the derivative of the velocity, so we find:

[tex]a(t)=v'(t)=3.40\cdot 10^5 -2(3.60\cdot 10^7)t=3.40\cdot 10^5 -7.20\cdot 10^7t[/tex]

Here we are told that the acceleration of the ball at the end of the barrel (so, when x = L, where L is the length of the barrel) is zero, so:

[tex]a(t')=0[/tex]

Where t' is the time at which the ball reaches the end of the barrel. Solving the last equation for t',

[tex]0=3.40\cdot 10^5 - 7.20\cdot 10^7t'\\t'=\frac{3.40\cdot 10^5}{7.20\cdot 10^7}=4.7\cdot 10^{-3} s[/tex]

b)

In part a, we have calculated that the time it takes for the cannonball to reach the end of the barrel is

[tex]t'=4.7\cdot 10^{-3} s[/tex]

We also know that the expression for the velocity of the ball is

[tex]v(t)=3.40\cdot 10^5 t-3.60\cdot 10^7 t^2[/tex]

Therefore, if we substituting t = t', we can find the velocity of the cannonball when it exits the barrel:

[tex]v(t')=3.40\cdot 10^5\cdot 4.7\cdot 10^{-3} -3.60\cdot 10^7 \cdot (4.7\cdot 10^{-3})^2=802.7 m/s[/tex]

And this is equal to the speed of the cannon: in fact, when it is in the barrel, the motion of the cannonball is along one direction only, this means that the speed is equal to the magnitude of the velocity.

c)

Here we want to find the length of the barrel.

From part a) and b), we know that the expression for the position of the cannonball at time t is

[tex]x(t)=(1.70\cdot 10^5)t^2 -(1.20\cdot 10^7)t^3[/tex]

Moreover, we also know that the ball exits the barrel at time of

[tex]t'=4.7\cdot 10^{-3} s[/tex]

This means that at t = t', the ball is at x = L, where L is the length of the barrel.

Therefore, we can find the length of the barrel by substituting the value of t' in the expression for x(t). Doing so, we find:

[tex]L=x(t')=(1.70\cdot 10^5)(4.7\cdot 10^{-3})^2-(1.20\cdot 10^7)(4.7\cdot 10^{-3})^3=2.51 m[/tex]

So, the length of the barrel is 2.51 m.

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