How to Calculate Acceleration in Physics 1-D Kinematics?

[Coder74]

Hi everyone, So far I have loved physics but I've been having trouble moving setting up the harder questions to solve for the answer. I would really appreciate the help! Thanks! :D

A car slows down from -27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m. What was it's acceleration?

\Delta x = 0.5(vf+vi)t - missing aKnown:
\Delta x -105m
vi - -27.7 m/s
vf - -10.9m/s
a - unkown

 

[MarkFL]

The kinematic formula you want is:

\(\displaystyle \overline{a}=\frac{v_f^2-v_i^2}{2\Delta x}\)

 

[Coder74]

Thanks so much, Mark! But how do you re-arrange the formula to what you want it to be?

 

[MarkFL]

Thanks so much, Mark! But how do you re-arrange the formula to what you want it to be?

I don't follow...it's already in the form we want to give us the acceleration. :D

 

[MarkFL]

Thanks so much, Mark! But how do you re-arrange the formula to what you want it to be?

Perhaps you meant how can we derive this formula. Let's begin with the definition of average acceleration:

\(\displaystyle \overline{a}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}\)

Now, for velocity that is changing linearly, the average velocity is:

\(\displaystyle \overline{v}=\frac{v_f+v_i}{2}\)

And average velocity is defined as:

\(\displaystyle \overline{v}=\frac{\Delta x}{\Delta t}\)

Hence:

\(\displaystyle 1=\frac{\overline{v}}{\overline{v}}=\frac{v_f+v_i}{2}\cdot\frac{\Delta t}{\Delta x}\)

Thus, we may state:

\(\displaystyle \overline{a}=\frac{v_f-v_i}{\Delta t}\cdot\frac{v_f+v_i}{2}\cdot\frac{\Delta t}{\Delta x}=\frac{v_f^2-v_i^2}{2\Delta x}\) :D

 

[Coder74]

Perhaps you meant how can we derive this formula. Let's begin with the definition of average acceleration:

\(\displaystyle \overline{a}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}\)

Now, for velocity that is changing linearly, the average velocity is:

\(\displaystyle \overline{v}=\frac{v_f+v_i}{2}\)

And average velocity is defined as:

\(\displaystyle \overline{v}=\frac{\Delta x}{\Delta t}\)

Hence:

\(\displaystyle 1=\frac{\overline{v}}{\overline{v}}=\frac{v_f+v_i}{2}\cdot\frac{\Delta t}{\Delta x}\)

Thus, we may state:

\(\displaystyle \overline{a}=\frac{v_f-v_i}{\Delta t}\cdot\frac{v_f+v_i}{2}\cdot\frac{\Delta t}{\Delta x}=\frac{v_f^2-v_i^2}{2\Delta x}\) :D

Thanks for the rapid reply! I really appreciate it..however.

I was lost in your explanation, it is truly in depth and an amazing point out.
But my teacher explained it in a completely different way. Here's an example of the "formulas" View attachment 6440

 

[topsquark]

Thanks for the rapid reply! I really appreciate it..however.

I was lost in your explanation, it is truly in depth and an amazing point out.
But my teacher explained it in a completely different way. Here's an example of the "formulas"

Actually it's the same. It's just that MarkFL ended up with a different form of an equation on your list. #4 in this case.

-Dan

 

[Coder74]

:c I don't get how they're the same. They look re-arranged to me. Physics isn't exactly my strongest subject and this is driving me crazy.. ;( I managed to find another practice question but since my teacher doesn't reply to her emails, I was wondering if you guys could help me out again... I really appreciate all of you guys <3!
View attachment 6441

 

[Euge]

Hi Coder74,

:c I don't get how they're the same.

In your formula sheet, Equation 4 is equivalent to Mark's formula. Starting from Equation 4

$$v_f^2 = v_i^2 + 2a\Delta x$$

subtract $v_i^2$ from both sides of the equation.

$$v_f^2 - v_i^2 = 2a\Delta x$$

Now divide both sides by $2\Delta x$.

$$\frac{v_f^2 - v_i^2}{2\Delta x} = a$$

Mark used $\bar{a}$ to represent acceleration. So his $\bar{a}$ is the same as your $a$.

I managed to find another practice question but since my teacher doesn't reply to her emails, I was wondering if you guys could help me out again... I really appreciate all of you guys <3!

Since the car comes to a stop in $15\; \text{m}$, its final speed $v_f$ is zero, and its displacement is $\Delta x = 15\; \text{m}$. Knowing that the car slows down at $a = -5.00\;\text{m/s$^2$}$, we can use Equation 5 in your formula sheet to solve for time $t$ required for the car to stop.

$$\Delta x = v_f t - \frac{1}{2}at^2$$

$$15.0 = 0 t - \frac{1}{2}(-5.00)t^2$$

$$15.0 = 2.50 t^2$$

$$6.00 = t^2$$

$$2.45 = t$$

Thus, it took $2.45$ seconds for the car to stop.