Question

1. in a heroic attempt to save the world from the flying eyes alien invaders, the thompsons (that odd couple from down the street) fires their 36m long super cannon at the approaching eyes. it is fired when the flying eyes are 1000m away and are approaching at 82m/s (top speed for a flying eye ya know). the thompsons fire a fresh happy giant spanish onion that experiences an acceleration of 109m/s² while in the cannon then continues at a constant velocity toward the eyes. how far from the cannons base will there be a lot of alien tears?

Explanation:

Step1: Find the muzzle velocity of the onion

We use the kinematic equation \( v^2 = u^2 + 2as \), where \( u = 0 \) (initial velocity in the cannon), \( a = 109 \, \text{m/s}^2 \), and \( s = 36 \, \text{m} \).
\[ v^2 = 0 + 2\times109\times36 \]
\[ v^2 = 2\times109\times36 = 7848 \]
\[ v = \sqrt{7848} \approx 88.59 \, \text{m/s} \]

Step2: Find the time taken for the onion to travel through the cannon

We use the equation \( v = u + at \), with \( u = 0 \), \( v \approx 88.59 \, \text{m/s} \), and \( a = 109 \, \text{m/s}^2 \).
\[ 88.59 = 0 + 109t \]
\[ t = \frac{88.59}{109} \approx 0.813 \, \text{s} \]

Step3: Find the distance the aliens travel during the cannon time

The aliens are moving at \( 82 \, \text{m/s} \), so distance \( d_1 = 82\times0.813 \approx 66.67 \, \text{m} \)

Step4: Find the remaining distance between the cannon and aliens after the cannon time

Initial distance is \( 1000 \, \text{m} \), so remaining distance \( D = 1000 - 66.67 - 36 = 897.33 \, \text{m} \) (subtracting the cannon length as the onion starts at the base and travels 36m in the cannon)

Step5: Find the time taken for the onion and aliens to meet after the cannon time

Let \( t' \) be this time. The onion's velocity is \( v \approx 88.59 \, \text{m/s} \), aliens' velocity is \( 82 \, \text{m/s} \).
\[ 88.59t' + 82t' = 897.33 \]
\[ 170.59t' = 897.33 \]
\[ t' = \frac{897.33}{170.59} \approx 5.26 \, \text{s} \]

Step6: Find the distance the onion travels after the cannon time

Distance \( d_2 = 88.59\times5.26 \approx 466.0 \, \text{m} \)

Step7: Find the total distance from the cannon base

Total distance \( D_{\text{total}} = 36 + 466.0 = 502.0 \, \text{m} \) (Wait, let's check again. Wait, in step 4, the remaining distance is between the onion's exit (36m from base) and the aliens. So when we calculate the time to meet after cannon, the onion travels \( v t' \) from the exit, and aliens travel \( 82 t' \) towards the exit. So the correct remaining distance is \( 1000 - 36 = 964 \, \text{m} \) (since the onion is 36m from base when it exits the cannon, and aliens are 1000m from base initially). Then the distance the aliens travel during cannon time is \( 82\times0.813 \approx 66.67 \, \text{m} \), so their distance from base when onion exits is \( 1000 - 66.67 = 933.33 \, \text{m} \), and onion is at 36m. So remaining distance between them is \( 933.33 - 36 = 897.33 \, \text{m} \) (correct as before). Then when they move towards each other, relative speed is \( 88.59 + 82 = 170.59 \, \text{m/s} \), time \( t' = 897.33 / 170.59 \approx 5.26 \, \text{s} \). Then onion travels \( 88.59\times5.26 \approx 466 \, \text{m} \) from exit, so total from base is \( 36 + 466 = 502 \, \text{m} \). Wait, but let's re - do step 4 and 5 correctly.

Correcting Step 4 and 5:

- Distance of aliens from base at time \( t \) (cannon time) : \( 1000 - 82\times t \), where \( t \approx 0.813 \, \text{s} \), so \( 1000 - 82\times0.813 \approx 1000 - 66.67 = 933.33 \, \text{m} \)
- Distance of onion from base at time \( t \): \( 36 \, \text{m} \) (since it traveled 36m in the cannon)
- The distance between onion (at 36m) and aliens (at 933.33m) is \( 933.33 - 36 = 897.33 \, \text{m} \)
- Now, the onion moves at \( v \approx 88.59 \, \text{m/s} \) towards the aliens, and aliens move at \( 82 \, \text{m/s} \) towards the onion. So their relative speed is \( 88.59 + 82 = 170.59 \, \text{m/s} \)
- Time to meet \( t'=\frac{897.33}{170.59}\approx5.26 \, \text{s} \)
- Distance onion travels after cannon: \( 88.59\times5.26\approx466 \, \text{m} \)
- Total distance from base: \( 36 + 466 = 502 \, \text{m} \)

Wait, but let's check with another approach. Let \( x \) be the distance from the cannon base where they meet.

- Distance traveled by onion: \( x = 36 + v\times t' \) (36m in cannon, then \( v t' \) after)
- Distance traveled by aliens: \( 1000 - x = 82\times(t + t') \), where \( t \) is time in cannon (\( \approx0.813 \, \text{s} \))

Substitute \( x = 36 + 88.59t' \) into the second equation:

\[ 1000-(36 + 88.59t')=82\times(0.813 + t') \]
\[ 964 - 88.59t' = 66.67+82t' \]
\[ 964 - 66.67=82t' + 88.59t' \]
\[ 897.33 = 170.59t' \]
\[ t'=\frac{897.33}{170.59}\approx5.26 \, \text{s} \]
Then \( x = 36+88.59\times5.26\approx36 + 466 = 502 \, \text{m} \)

Answer:

The distance from the cannon's base is approximately \(\boxed{502}\) meters (or more accurately, re - checking the velocity calculation:

\( v^2 = 2as = 2\times109\times36 = 7848 \), \( v=\sqrt{7848}\approx88.59 \, \text{m/s} \) (correct)

Time in cannon: \( t=\frac{v}{a}=\frac{88.59}{109}\approx0.813 \, \text{s} \) (correct)

Distance aliens travel in cannon time: \( 82\times0.813\approx66.67 \, \text{m} \) (correct)

Distance between onion (at 36m from base) and aliens (initial 1000m from base) after cannon time: \( 1000 - 36 - 66.67 = 897.33 \, \text{m} \) (correct)

Relative speed: \( 88.59 + 82 = 170.59 \, \text{m/s} \) (correct)

Time to meet after cannon: \( \frac{897.33}{170.59}\approx5.26 \, \text{s} \) (correct)

Distance onion travels after cannon: \( 88.59\times5.26\approx466 \, \text{m} \) (correct)

Total distance: \( 36 + 466 = 502 \, \text{m} \) (correct)

So the final answer is approximately \(\boxed{502}\) meters.