Understanding the math in physics

[benshields]

I just started learning kinematics in my college physics class and I am pretty thrilled. Question is am I supposed to be making sense of the mathematics in the equations? Because I feel like I'm just regurgitating formulas and I don't really think that's what science is about.

For example, one of the kinematic equations is D=Vi(t)+half of(a)(t)squared.
I want to understand why the time is squared at the end and why is it even divided by one half in the first place?

 

[olivermsun]

Think of the graph of your speed if you start from 0 and accelerate at a constant rate.

The graph of $V(t)$ will be a straight line with constant slope a: $V(t) = at$.

The triangular area under the graph is the distance you've traveled.

The area of a right triangle is $\frac{1}{2}$(base) * (height) = $\frac{1}{2} t * at= \frac{1}{2}a t^2$.

 

[benshields]

That makes much more sense. I just wasn't thinking about it graphically. Thank you so much man!

 

[Fredrik]

Do you understand derivatives and integrals? Let x be the function such that for all times t, x(t) is the object's position at time t. Suppose that we would like to know how an object moves under the influence of a force that doesn't change with time. Since force equals mass times acceleration, to assume that the force is constant is to assume that the acceleration is constant. To say that the acceleration is constant is to say that there's a real number $a$ such that $x''(t)=a$ for all t. Integrate this, and we find that for all t, we have $x'(t)=at+C$, where C is a constant that can be determined by setting t=0. We have $x'(0)=a 0+C=C$. So C is the velocity at time 0. We therefore choose to denote it by $v_0$. In this notation, we have $x'(t)=at+v_0$ for all t. Now integrate this, and we find that for all t, we have $x(t)=\frac 1 2 a t^2+v_0 t+C$, where C is another constant. This implies that $x(0)=\frac 1 2 a 0^2+v_0 0+C=C$. So C is the position at time 0. We can make it 0, simply by choosing the coordinate system so that x(0)=0, i.e. so that the object is at position coordinate 0 at time 0. This gives us the final result $x(t)=\frac 1 2 at^2+v_0t$ for all t.

 

[HalfLostAndSearching]

As a scientist, it is important to not only understand the concepts and principles in physics, but also the mathematical equations that are used to represent them. While it may seem like you are just regurgitating formulas, it is crucial to be able to manipulate and interpret these equations in order to make meaningful connections and draw conclusions about the physical world.

In regards to the specific kinematic equation you mentioned, the time being squared at the end is a result of the mathematical derivation of the equation. In kinematics, we are often interested in the relationship between position, velocity, acceleration, and time. The equation you mentioned, D=Vi(t)+1/2(a)(t^2), is a representation of this relationship, where D is the final position, Vi is the initial velocity, a is the acceleration, and t is the time. The time being squared at the end is a result of the mathematical manipulation of the equation to solve for the final position. It is not something that can be easily explained without delving into the mathematical derivation of the equation.

As for the division by one half, this is a result of the mathematical relationship between acceleration and time. In this equation, the acceleration is multiplied by the square of time, which results in a unit of distance. However, in order to represent this distance accurately, it must be divided by one half to account for the fact that the acceleration is acting over a period of time.

In conclusion, while it is important to understand the concepts and principles in physics, it is equally important to understand the mathematical equations that represent them. This will allow you to make meaningful connections and draw conclusions about the physical world. Keep asking questions and seeking to understand the reasoning behind the equations, and you will continue to develop a deeper understanding of physics.