# Understanding the math in physics

1. Sep 12, 2014

### 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?

2. Sep 12, 2014

### 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) = $\frac{1}{2} t * at= \frac{1}{2}a t^2$.

3. Sep 12, 2014

### benshields

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

4. Sep 12, 2014

### Fredrik

Staff Emeritus
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.