__Key Concepts:__**Displacement**
**Velocity **
**Acceleration **
**Equations of motion**
**Projectile motion**
QUESTION 1:
"A car is driving down a highway. The attached graph shows its **displacement** as a function of time. Determine its **speed** and its equation of motion."

How do we figure out the speed of the car just by looking at this graph? It's actually pretty simple. Think about it like this: If it takes me 1 hour to drive between two cities that are 100 km away from each other, then clearly my speed will have been:

So basically we need 2 things from the graph: **How far the car has gone **and **how much time it spent going this far**. To do this, we look at how much the graph has "**gone up**" compared to how much it has "**gone forward**".:

Very nice. To determine the equation of motion, we write:

QUESTION 2:
"A car is driving down a highway. The attached graph shows its **speed** as a function of time. Determine its acceleration and its **equation of motion.**"

Ok, so this time the task is a little different. We need to find the "**acceleration**" of the car. What is "acceleration" anyways? Let's make a small drawing:

Basically, we can say that **the car is accelerating because it speed keeps changing**. In this case the speed keeps **increasing** so the acceleration must be **positive.**
We can define acceleration as:

The acceleration is 3m/s^2. This means that for every second that passes, the speed increases by 3m/s.
We still need the **"Equation of motion"**. How do we find that?
In the first question, we found that

Basically we can find the speed by using calculus to take the **derivative** of the distance. We can do the same thing for the acceleration:

Here is the point:
**"If we have x(t), we can find v(t) and a(t) by simply taking derivatives!"**
Ok, but in this situation, we have the a(t)=3m/s^2. We can find x(t) by doing the **opposite of taking the derivative: Taking the integral!**

Let's try to graph it:

Weird. The displacement graph is sort of bending upwards. I mean that kind of makes sense. The car is **accelerating** so it's going faster and faster. In the first 4 seconds, the car is not going very fast, so it only covers 25m. In the next 4 seconds it covers 75m.
We can write the **General equation of motion:**

^This works **if acceleration is constant**.
QUESTION 3:
"A ball is thrown sideways at a speed of 3m/s from a 100m tall building. How far from the base of the building does it hit the ground?"
Let's make a sketch to figure out what is going on:

In the beginning, the ball is just flying straight to the right because of its initial speed. However, gravity pulls it down and gives it a vertical speed. After a certain amount of time, it hits the ground a distance d away from the building.
Here is a useful trick we can use:
**"We can write down 2 equations of motion. One for motion in the horizontal x-direction and one for the vertical y-direction."**
How should we do this for the current case?
The initial speed of the ball in the x-direction is 3m/s. No force is pulling the ball to the right, so there is no acceleration.
For the y-direction, the ball starts at y=h and gets pulled down by gravity. The acceleration it experiences is a=-9.82 m/s^2. So in total we can write:

Very nice. This is a good start. We need to find the distance at which the ball hits the ground. We know that it hits the ground after a certain number of seconds. Let us write:

Here "ti" is the "impact time". We already know v0 and we want to find d. So can we find an expression for ti somehow? Take a look at the y-equation. When y=0 we know that the ball has hit the ground. We can therefore write:

^So the impact time depends on how far up we start and how much the ball accelerates due to gravity. Let's plug this into the expression for the impact distance:

The ball hits the ground 13.4m away from the base of the building.

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