# Acceleration

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Newton's 2nd Law, Acceleration, With Friction, Net Force Above the Horizontal

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1 of 6 videos by Brian Swarthout

## Key Questions

• Acceleration is the derivative of velocity with respect to time. In simpler terms (for an algebra-based physics approach), acceleration is also a change in velocity divided by a change in time.

Generally, when finding acceleration in 1D motion problems you are going to want to apply one of the kinematic equations for constant acceleration motion. To use these equations, I recommend writing down EVERYTHING that you know from a given problem and play with the equations until you find something that you can calculate, based on which variables you know.

All the equations depend on the following variables:

$\Delta x$ = change in position
${v}_{0}$ = initial velocity
${v}_{f}$ = final velocity
a = acceleration
$\Delta t$ = change in time

And now, the KINEMATIC EQUATIONS!

1) $\Delta x = \frac{{v}_{o} + {v}_{f}}{2} \cdot \Delta t$

2) $\Delta x = {v}_{o} \cdot \Delta t + \frac{1}{2} \cdot a \cdot \Delta {t}^{2}$

3) ${v}_{f}^{2} = {v}_{o}^{2} + 2 \cdot a \cdot \Delta x$

4) ${v}_{f} = {v}_{o} + a \cdot \Delta t$

We can also find acceleration using Newton's 2nd law, âˆ‘F=mâ‹…a.

If the net force and the mass is known, $a = \frac{\sum F}{m}$

• let ${F}_{1}$ = initial acceleration and ${F}_{2}$ = final acceleration and t be the time taken for this acceleration to occur. then
Avg Acceleration= $\frac{{F}_{2} - {F}_{1}}{t}$
You can replace terms with standard motion equations to get a more elegant equation.

• Let's use Newton's Laws of Motion to determine a few things...

First, the acceleration of the car:
$a = \frac{{v}_{f} - {v}_{i}}{t}$, where v_f = 26.82 m/s; v_i = 0; t = 8.0 s

$a = \frac{26.82 - 0}{8.0} = 3.35 \frac{m}{s} ^ 2$

Now, to determine the final speed when v_i = 22.35 m/s; t = 5 s; a = 3.35 m/s^2

${v}_{f} = {v}_{i} + a t$
${v}_{f} = 22.35 + 3.35 \cdot 5 = 39.1 \frac{m}{s}$

The same problem in English units of miles per hour:

Acceleration of the car from 0mi/h to 60mi/h in 8.0s:

$a = \frac{{v}_{f} - {v}_{i}}{t}$, where v_f = 60mi"/"h; v_i = 0; t = 8.0 s

a = (60mi"/"h - 0)/"8.0s" = 7.5mi"/"h"/"s

Final speed when initial velocity is 50mi/h:

${v}_{i} = 50 m i \text{/} h$; $a = 7.5 m i \text{/"h"/} s$; $t = 5.0 s$

${v}_{f} = {v}_{i} + a t$

${v}_{f} = 50 m i \text{/} h$ + ($7.5 m i \text{/"h"/} s$)($5.0 s$)

${v}_{f} = 50 m i \text{/"h + 37.5mi"/"h = 87.5mi"/} h$

## Questions

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