Question

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 ✕ 10⁵ m/s²)t² − (1.20 ✕ 10⁷ m/s³)t³.

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

At what time is the cannonball at the end of the cannon? m

Answer 1

From vC)and a we get Veloch and accelerakon dt Junckons -( 3.6x10 3.4 × 10 Given 4hat the ace elerahm as iㅏ eisbみe cannon is zero a) So en a(+) =o gives us the time ,when acceler a Kon is 2e, ,ash; s the hme faken to tverel he leath o sah barel because ot the 7-2 ×○ラ 4-72 x103Sec りSo now we put this t-4.72x103 sec inve) t Set the veloaty So it Velocity is VCt) - 802.78 se c) Put t= 4.72 xos Sec in 2(t); we get the dist ance ta,elle t-o to t=472x, i-ethe length ol barrel! orn 3 e) lengh of bavreli | L = 2.525 m