1D Graphs, Kinematics, & Calculus

GraphsPosition Velocity Acceleration

ConstantPosition

ConstantVelocity

ConstantAccel.

GraphsPosition Velocity Acceleration

ConstantPosition

ConstantAccel.

ConstantVelocity

GraphsPosition Velocity Acceleration

ConstantVelocity

y = mx + bx (x/t) t xi= +v

Notice the slope of the x vs. t graph is the velocity!

f

With no acceleration, distance = speed * time!

x = vt

GraphsPosition Velocity Acceleration

ConstantPosition

ConstantVelocity

ConstantAccel.

GraphsPosition Velocity Acceleration

ConstantAccel.

y = mx + bv (v/t)t vi= +a

f

Notice the slope of the v vs. t graph is the acceleration!

y = ax2 + bx + cx xi+= t2 t½a +vi

fvf = vi + atx = vit + ½at2

Key Ideas

• Slope of x vs. t graph is the velocity.

• Slope of v vs. t graph is the acceleration.

Position

x

t

Δt

Δx

Δ delta always final minus initial

Since this is from the beginning to the end, it gives us the average velocity.

What if we want to find the velocity at a specific point (instantaneous velocity)? What slope is this going to be?

Position

x

t

As the time approaches 0, the slope becomes the tangent line and the velocity becomes the velocity at that instance. This is all a derivative is!The slope of a tangent line.

What do you think the derivative of the velocity vs. time graph is?

Velocity a?

Key Ideas

• Slope of x vs. t graph is the velocity.– The derivative of the x vs. t graph is the acceleration

• Slope of v vs. t graph is the acceleration.– The derivative of the v vs. t graph is the acceleration.

Basic Derivatives (How to)

v = 2t = 2(2) = 4the value of the exponent multiplies the coefficient

the exponent drops by a power of 1

ex. A ball’s position is described by the following equation x = t2 find the ball’s velocity at 2 seconds.

PSA: the derivative of a constant is 0

You Try1. Find the derivative of the following equation:

x = 4t3 + ½t2 + 6t + 4x = 4

x = 12t2 + t + 6

t3 + ½ t2 + 6t + 42 1

2. A book falls from the ceiling with the following equation describing its position. x = -4.9t2 - 3t + 2a. Find the book’s position at 0.653 seconds.b. Find the book’s velocity at 0.653 seconds.c. Find the book’s acceleration at 0.653 seconds.

a. x = -4.9t2 – 3t + 2 x = -4.9(0.653)2 – 3(0.653) + 2 = -2.05 mb. remember the velocity is the derivative of the

position equation

c. remember the velocity is the derivative of the position equation

One More Thing to look at…If the slope (y/x) means something, then surely y*x (area under the line) must mean something!

ConstantVelocity

? = y * xv t= *x

ConstantAccel.

? = y * xa t= *v

How about the acceleration graph?

What if the graph is oddly shaped?How would you find the area under the curve?

This is an integral! (and in its most basic form is opposite of a derivative)

Key Ideas• x vs. t graph

– Slope (derivative) is the velocity.• v vs. t graph

– Slope (derivative) is the acceleration.– The Integral (Area Under the Curve) is the

displacement.• a vs. t

– The Integral (Area Under the Curve) is the velocity.

Two parts to a graph: y/x Slope (derivative) y*x Area Under Curve (Integral)