# Calculating the Displacement (Having Constant Acceleration)

• epuen23
In summary, the conversation discusses a student's struggle with understanding how to correctly use formulas and the correct order of operations when solving problems. The student learns that the exponentiation should be done first, followed by the multiplication by other quantities. Additionally, it is important to pay attention to units in order to obtain the correct result. The student expresses gratitude for the help and realizes that their mistake was a simple one.
epuen23
Homework Statement
A pinewood derby car starts down a constant slope with a constant acceleration of 2.2 m/s2. What will be its displacement after 2.4 s?
Relevant Equations
d=vt+1/2at^2
I'm just having trouble understanding 1) how to plug in the formulas correctly and 2) the correct manner of attacking this.

I feel like I'm missing something basic/simple. Any help is greatly appreciated! So far I've got:

D = (0) + .5(2.2m/s^2 x 2.4s)^2

Enrique

epuen23 said:
D = (0) + .5(2.2m/s^2 x 2.4s)^2

No, ##a## is not squared, just the time ##t##. ##at^2## means that ##a## is multiplied by ##t^2##.

This has to do with what is called "order of operations". Sometimes students are taught the mnemonic PEMDAS to memorize that order: Parentheses, Exponentiation, Multiplication/Division, Addition/Subtraction.
So without any parentheses you would do the exponentation ##t^2## first, before the multiplication by the other quantities.

In addition to #2, note that your units do not come out correctly if you do (at)^2 rather than at^2. Instead of meters, you would get (m/s)^2, which is not a good unit for a displacement.

RPinPA said:
No, ##a## is not squared, just the time ##t##. ##at^2## means that ##a## is multiplied by ##t^2##.

This has to do with what is called "order of operations". Sometimes students are taught the mnemonic PEMDAS to memorize that order: Parentheses, Exponentiation, Multiplication/Division, Addition/Subtraction.
So without any parentheses you would do the exponentation ##t^2## first, before the multiplication by the other quantities.

Oh. My. Gosh! Thank you so much. I don't know how I missed that. Thank you!

Orodruin said:
In addition to #2, note that your units do not come out correctly if you do (at)^2 rather than at^2. Instead of meters, you would get (m/s)^2, which is not a good unit for a displacement.

Orodruin, thank you so much. I guess it was just a silly mistake I made and didn't catch it. Thank you!

## 1. What is displacement?

Displacement is the distance and direction of an object's change in position from its starting point to its ending point. It is a vector quantity, meaning it has both magnitude (size or numerical value) and direction.

## 2. How is displacement calculated?

Displacement is calculated by subtracting the initial position from the final position of an object. This can be represented by the equation: Δx = xf - xi, where Δx is displacement, xf is final position, and xi is initial position.

## 3. What is constant acceleration?

Constant acceleration is when an object's velocity changes at a constant rate over time. This means that the object's speed increases or decreases by the same amount in each unit of time.

## 4. How do you calculate displacement with constant acceleration?

To calculate displacement with constant acceleration, you can use the equation: Δx = viΔt + 1/2a(Δt)^2, where Δx is displacement, vi is initial velocity, a is acceleration, and Δt is the change in time.

## 5. Can displacement be negative?

Yes, displacement can be negative. This means that an object has moved in the opposite direction from its starting point. Positive displacement indicates movement in the same direction as the initial position, while negative displacement indicates movement in the opposite direction.

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