Calculating the Displacement (Having Constant Acceleration)

[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
Answer: 6.3 m

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

 

[RPinPA]

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.

 

[Orodruin]

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.

 

[epuen23]

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!

 

[epuen23]

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!