MHB How to Calculate Acceleration in Physics 1-D Kinematics?

AI Thread Summary
The discussion focuses on calculating acceleration using kinematic equations in one-dimensional motion. A user presents a problem involving a car's deceleration and seeks clarification on rearranging the kinematic formula for acceleration. Another participant explains the derivation of the acceleration formula, demonstrating its equivalence to a different form found in the user's formula sheet. The conversation highlights the importance of understanding various representations of the same equations in physics. Ultimately, the user expresses gratitude for the assistance while seeking further help with additional practice questions.
Coder74
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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
 
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The kinematic formula you want is:

$$\overline{a}=\frac{v_f^2-v_i^2}{2\Delta x}$$
 
Thanks so much, Mark! But how do you re-arrange the formula to what you want it to be?
 
Coder74 said:
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
 
Coder74 said:
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:

$$\overline{a}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}$$

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

$$\overline{v}=\frac{v_f+v_i}{2}$$

And average velocity is defined as:

$$\overline{v}=\frac{\Delta x}{\Delta t}$$

Hence:

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

Thus, we may state:

$$\overline{a}=\frac{v_f-v_i}{\Delta t}\cdot\frac{v_f+v_i}{2}\cdot\frac{\Delta t}{\Delta x}=\frac{v_f^2-v_i^2}{2\Delta x}$$ :D
 
MarkFL said:
Perhaps you meant how can we derive this formula. Let's begin with the definition of average acceleration:

$$\overline{a}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}$$

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

$$\overline{v}=\frac{v_f+v_i}{2}$$

And average velocity is defined as:

$$\overline{v}=\frac{\Delta x}{\Delta t}$$

Hence:

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

Thus, we may state:

$$\overline{a}=\frac{v_f-v_i}{\Delta t}\cdot\frac{v_f+v_i}{2}\cdot\frac{\Delta t}{\Delta x}=\frac{v_f^2-v_i^2}{2\Delta x}$$ :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
 

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Coder74 said:
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
 
: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
 

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Hi Coder74,

Coder74 said:
: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 $a$.
Coder74 said:
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.
 

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