Calculus In The REAL World!!

[@justplay2508]

Its a kinematic equation 10t is simply V(initial)*t + 50 is Y(initial) -1/2gt^2 (g is acceleration due to gravity which is -9.8)

[@mihailmilev9909]

Thanks

[@mihailmilev9909]

21 like

[@paranoyd1]

y(t) = y0 + v0t + (1/2)at^2 where y0 is the initial position (50 in this case), v0 is the inicial velocity (10), and a is the acceleration (-g in this case)

[@mihailmilev9909]

5th like

[@Lecommandant_camroun]

Yup Anyways remember Jesus loves you so he died for you because he wants to know you❤Repent, God bless❤ Know then in your heart that as a man disciplines his son, so the Lord your God disciplines you. Deuteronomy 8:5

[@tunistick8044]

our exams be like "calculate α the angle between Vi and the horizontal plane and β the angle between Vf and the vertical plane"

[@NotYuji_weeb]

I was always bad at math but i learn something good always from you

[@mihailmilev9909]

7th like

[@NotYuji_weeb]

@@mihailmilev9909 cool

[@Eduardo-tq5sk]

Thank you profe !

[@draculq3031]

-4.9t^2 represents gravity in meters 50 is the starting position so 10t is the initial velocity just by looking at the equation

[@The1200r]

Uh huh, yep. That's what I was thinking.

[@NoosaHeads]

Isn't it a question of dy/dt being slope "0" and wouldn't that be a 0= -9.8t +10 giving 9.8t =10 ie T= 10/9.8 ie approx 1.1 seconds?

[@BananaDude508]

OOOOH OK THAT MAKES SENSE NOW

[@jazzle9121]

so the ball is moving 10 meters per second even time is still zero? doesn't make sense to me. How can the ball have a velocity when it is at rest.

[@obinnanwakwue5735]

t = 0 does not mean the ball is at rest. v = 0 means the ball is at rest. t = 0 means the beginning of the timeframe.

[@januszek1760]

@@obinnanwakwue5735 But on the picture ball is at rest for t=0, it's confusing.

[@irastazen6777]

Imagine that you set up your phone on the side of a road/street in order to record a car passing by. Let's say the car is at a constant 20m/s. Then you watch the video that you just recorded of that car passing by at 20m/s. You could pause the video at any given moment and the car would be fixed in place for you, but you know that when you play it again, the car is gonna move at 20m/s. Time = 0 but initial velocity is 20m/s

[@tunistick8044]

who told you that we didn't throw it already before t=0?

[@DrDeuteron]

@@januszek1760 things emitting blue arrows are not at rest

[@rossfriedman6570]

How does it start at a non zero velocity if it's falling off a ledge?

[@DrDeuteron]

by definition

[@cppghost]

It's going up initially, then stopping and falling down

[@thomasolson7447]

This graph is a parabola that is slightly different from what you see in the video. It goes from -infinity to infinity on the x-axis, which represents time. It first crosses the x-axis at (50-30*sqrt(30))/49 or approximately -2.3329952500316 seconds. It crosses the y-axis at 10 m/s which is the tangent (rate of change) at 0 time. The y(0) is -4.9*(0)^2+10*(0)+50=50 meters. At 50 m above the x-axis, it has an initial velocity of 10m/s and velocity decrease until 50/49 (-9.8*t+10=0) or approximately 1.0204081632653 seconds. It increases in position on the y-axis till 50/49 seconds , then the position decreases. So the y-axis position at y(50/49) = -4.9*(50/49)^2+10*(50/49)+50=(2700/49) or approximately 55.1020408163265 meters. It crosses the x-axis a second time at position t=(30*sqrt(30)+50)/49 or approximately 4.3738115765622 m. The velocity at that time is -6*sqrt(30) m/s or approximately -32.86335345031 m/s. The graph suggests there is an initial force being applied and the initial velocity is 0, but that is wrong. The true graph is a full parabola from -infinity to infinity. There might be a way to figure out what kind of force you would need to apply at t=0, but you would need a mass on whatever is being thrown.

[@highgglsgf]

kinematics simplified

[@Bellaisbald]

Is more simple to say 10T is = 10 meters per second 😅

[@tahliameadows4767]

How is the velocity function equivalent to the derivative of the original y function that was given befor you found it? I dont understand

[@Lecommandant_camroun]

because v = dr/dt v = velocity r = position t = time (Derivative of position with respect to time) Anyways remember Jesus loves you so he died for you because he wants to know you❤Repent, God bless❤ Know then in your heart that as a man disciplines his son, so the Lord your God disciplines you. Deuteronomy 8:5

[@beinganangeltreon]

ok, this is kind of comical, but what could you ever do to a moving object that would take a negative velocity or time (film backwards), and then multiply it with another negative, to get an imaginary number, one possibility is towards the screen could be positive, and deeper into the screen as negative, if the entire moving thing were receding into the screen then it would be -4.9t if we treat 4.9T^2 as negative as its towards earth, oops, t^2 is automatically 4.9i from the negative exponent as a ^2, but the thing is, if you derive that 4.9i you either get -4.9 or there's some other weird thing about removing a unit amount from the exponent, so i just goes to zero, but if you do that you still have just regular 4.9, or maybe you have -4.9, but if you have -4.9 then gravity(9.8) is half force, or its only 4.9t, and goes half velocity, but if either of those is true then, somewhat incredibly making Y=0 then I still get rid of that 4.9 and still have 10 as the velocity. and, that's what not passing calculus but getting the occasional test question right. I really like math though, it's fun, I'm just technically ignorant. Study math!

[@beinganangeltreon]

There is a solution with moving forward or at depth into the screen though, because with some equation you get the equivalent of an orbital circle velocity, or assuming a plane, that velocity that keeps the thing traversing screenward at a nonfinite but continuous nearness to the plane that never meets the surface of the plane. I think that might be nonfinity, and of course orbits work because the sphere has a curvature. entertaining.

[@cppghost]

Wtf are you talking about?(this is very dumb)

[@backwashjoe7864]

Where did you get the units of m/s at the end? There were no units of distance specified up to that point.

[@draculq3031]

4.9t^2 represents gravity in meters, -16t^2 is feet

[@backwashjoe7864]

​@@draculq3031 Not if you don't state the units explicitly.

[@draculq3031]

@@backwashjoe7864 it's pretty much implied from the figure -4.9t^2, but yeah he didnt state it explicity. When ur dealing with problems like this, where its an object moving along like a parabolic path(i.e something shot from like a cannon) the -4.9t^2 or -16t^2 will give you the units indirectly

[@DrDeuteron]

S.I. units are implied, but that's how you crash orbiters into Mars.

[@Nxck2440]

This isn't real world, you need to find that y(t) from geometry or sth

[@thelifeofibo]

No you don’t need to find y(t) in most questions???

[@mihailmilev9909]

Nice

[@Iamrich00]

How did u know that you have to take the derivative to find velocity

[@ChrisJab4]

Derivative is the slope and the slope of a position-time graph is change in distance/change in time, which we call velocity. Also you could just remember that the derivative of position is velocity and derivative of velocity is acceleration if what I said before doesn't make sense

[@mynak4291]

From going to position to velocity or velocity to acceleration We do differentiation and to do the reverse we integrate

[@ChrisJab4]

@Px Coffee Yes that's correct. What I was saying is that the velocity of the graph can be found with the derivative but it is also the slope of the tangent line

[@DrDeuteron]

its the definition of velocity.

[@daco-shitpost]

just use x = x0 + v0t + at^2 / 2, the coeficient next to t^2 is -g/2, the coeficient near t is v0, the remaining coeficient is x0 (height)

[@cppghost]

Yes, but if I gave you x = arctan²(x³) - e²sin(cos(x)), you wouldn't be able to find the velocity, would you? 😊

[@daco-shitpost]

​@@cppghost Actually, I would. Derivative of that, using the chain rule 3 times, is: 2 arctan(x³)•1/(1+x^6)•3x² - e²•cos(cos(x))•(-sin x) Which when substituting 0 gives: arctan(0)•...+sin(0)•...=0 And where would you find that kind of formula anyways? The method I described can be used only with motion with constant acceleration, but it is simpler

[@Lecommandant_camroun]

It was technically be y because x is horizontal position 😊😊😊😊😊☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁 Anyways remember Jesus loves you so he died for you because he wants to know you❤Repent, God bless❤ Know then in your heart that as a man disciplines his son, so the Lord your God disciplines you. Deuteronomy 8:5

[@hardani.ismunabil]

BALL

[@karunanidhanpandey8689]

10sin therta will be the ans

[@badretdyn]

производная

[@monak4854]

Why not 50?

[@monak4854]

When x is zero. Then y is 50. Right. 😅

[@masterdementer]

Well it's initial position will be at the coordinate of y=50. That won't give the velocity. You need to first take the derivative of the position to get the equation for velocity. And when you put t=0 in that equation you will get initial velocity. You can even find the acceleration of you take another derivative of the velocity equation and that will give you a constant of 9.8 (which lines up perfectly with the diagram as the object is free falling and acceleration is just due to gravity g=9.8)

[@Lecommandant_camroun]

This is just y_final (or y(t))= y_initial + v_initial*t + at^2/2 y_initial is 50m y_final (at t 0) is always the same as y_inital, so it's 50 m t is 0 a is just g, so -9.8 m/s/s v_inital is 10 m/s Anyways remember Jesus loves you so he died for you because he wants to know you❤Repent, God bless❤ Know then in your heart that as a man disciplines his son, so the Lord your God disciplines you. Deuteronomy 8:5