It’s been a while, but I think it’s quite trivial.
After one second, they span a right angled triangle, therefore (using a² + b² = c²) their distance is √(5²+1²) = ~5.1 ft
They move at constant speed, therefore they seperate at 5.1 ft/s. That means at 5s it’s just 5.1 × 5 = 25.5 ft for the distance and their speed is still the same.
They each move at a constant speed, but the distance between them doesn’t increase at a constant pace. See my other comment.Edit: I am dumb, and looked at the wrong number.
I’m trying to apply the most simple math possible and it seems to add up.
After one second, their distance is √(5² + 1²) = ~5.1 ft
After two seconds, their distance is √(10² + 2²) = ~10.2 ft
After three seconds, it’s √(15² + 3²) = ~15.3 ft
As speed is the rate of change of distance over time, you can see it’s a constant 5.1 ft/s. You’re free to point out any error, but I don’t think you need anything more than Pythagoras’ theorem.
The question specifically asks for their seperation speed at 5s to ignore any initial change in their speed as they first need to accelerate, I’d assume.
Ah sorry, I’m tired and made a mistake. I quickly made a spreadsheet (because keeping track of numbers is hard), and I was looking at the wrong column in the sheet. My bad!
You were tired so you made a spreadsheet to calculate the differential equation quiz from a meme?
Yes, compared to doing the calculations in my head lol
I work in mysterious ways
I don’t see why the distance between them isn’t growing at a constant speed.
At any given time t seconds after separation, the boy is 5t north, and the girl is 1t east. The distance between them is defined by the square root of ((5t)^2 + (t)^2 ), or about 5.099t.
In other words, the distance between them is simply a function defined as 5.099t, whose first derivative with respect to time is just 5.099.
Depends on where they met each other. If they for example fell in love during the main event of a trip to the north pole, that would change things a lot.
there is no north at the north pole so actually that’s the one place it can’t be
If you’re at the south pole, would every direction count as north?
Sure, but there is a north say 30 ft away from the north pole.
Its pretty convenient that its raining, which means you can ignore the coefficient of friction since the surface is slippery
It’s* pretty convenient that it’s* raining
Differential calculus? That looks more like algebra. Their speed is constant.
I agree, it is not calculus, it’s trigonometry.
Each of their speeds is constant, but different, and they’re walking in different directions.
Their distance is the hypotenuse of a triangle with sides 5t and t which will be root((5t)2 + t2). So the distance at time t of the ex lovers will be root(26) × t. You can basically grasp intuitively that the speed is indeed constant and equals to the root(26)=5.1 ft/sec. Technically you’d use the derivative power rule to drop the t and get the speed.
Look. Teachers have some unresolved shit as well.
reminds me of that one song, proof that geometric construction can solve all love affairs or something like that
what
Who hurt the math teacher?
9,14 Meter
The question states “how fast”, not “how far”, thus you need to give the acceleration at that moment.
At t=0, the boy and girl both haven’t moved, so their positions are 0. The distance between them is also 0, as is their acceleration.
The boy’s distance in meters is t*1.524, the girl’s distance is t*0.3048. The distance between them is sqrt( b^2 * g^2 ). The velocity is the current distance minus the previous distance.
At t=1, b=1.524m, g=0.305, d=sqrt( g^2 * g^2 )=0.465, v=d-d^(t-1)=0.465m/s.
At t=5, b=7.62, g=1.524, d=11.613, and v=4.181m/s.
Edit: fixed markdown
