Question

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osuniphys1 3.3.wa.032.tutorial.
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
$x(t) = (1.50 \times 10^{5}\\ \text{m/s}^2)t^2 - (1.80 \times 10^{7}\\ \text{m/s}^3)t^3$.
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
\\(\boldsymbol{\square}\\) s
(b) speed of a cannonball as it exits the barrel
\\(\boldsymbol{\square}\\) m/s
(c) length of the barrel
\\(\boldsymbol{\square}\\) m

Explanation:

Step1: Find acceleration function

Acceleration is the 2nd derivative of position.
First, find velocity (1st derivative of $x(t)$):
$v(t) = \frac{dx}{dt} = 2(1.50 \times 10^5)t - 3(1.80 \times 10^7)t^2 = 3.00 \times 10^5 t - 5.40 \times 10^7 t^2$
Then find acceleration (derivative of $v(t)$):
$a(t) = \frac{dv}{dt} = 3.00 \times 10^5 - 1.08 \times 10^8 t$
Set $a(t)=0$ (acceleration is zero at exit):
$0 = 3.00 \times 10^5 - 1.08 \times 10^8 t$

Step2: Solve for exit time $t$

Rearrange to solve for $t$:
$t = \frac{3.00 \times 10^5}{1.08 \times 10^8}$
$t = 2.777... \times 10^{-3} \approx 2.78 \times 10^{-3}$ s

Step3: Calculate exit speed

Substitute $t$ into $v(t)$:
$v(2.78 \times 10^{-3}) = 3.00 \times 10^5(2.78 \times 10^{-3}) - 5.40 \times 10^7(2.78 \times 10^{-3})^2$
First term: $3.00 \times 10^5 \times 2.78 \times 10^{-3} = 834$
Second term: $5.40 \times 10^7 \times (7.7284 \times 10^{-6}) = 417.3336$
$v = 834 - 417.3336 \approx 417$ m/s

Step4: Calculate barrel length

Substitute $t$ into $x(t)$:
$x(2.78 \times 10^{-3}) = 1.50 \times 10^5(2.78 \times 10^{-3})^2 - 1.80 \times 10^7(2.78 \times 10^{-3})^3$
First term: $1.50 \times 10^5 \times 7.7284 \times 10^{-6} = 1.15926$
Second term: $1.80 \times 10^7 \times 2.1485 \times 10^{-8} = 0.38673$
$x = 1.15926 - 0.38673 \approx 0.772$ m

Answer:

(a) $2.78 \times 10^{-3}$ s
(b) $417$ m/s
(c) $0.772$ m