# 3.3 Average and instantaneous acceleration  (Page 2/7)

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The term deceleration can cause confusion in our analysis because it is not a vector and it does not point to a specific direction with respect to a coordinate system, so we do not use it. Acceleration is a vector, so we must choose the appropriate sign for it in our chosen coordinate system. In the case of the train in [link] , acceleration is in the negative direction in the chosen coordinate system , so we say the train is undergoing negative acceleration.

If an object in motion has a velocity in the positive direction with respect to a chosen origin and it acquires a constant negative acceleration, the object eventually comes to a rest and reverses direction. If we wait long enough, the object passes through the origin going in the opposite direction. This is illustrated in [link] .

## Calculating average acceleration: a racehorse leaves the gate

A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?

## Strategy

First we draw a sketch and assign a coordinate system to the problem [link] . This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.

We can solve this problem by identifying $\text{Δ}v\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{Δ}t$ from the given information, and then calculating the average acceleration directly from the equation $\stackrel{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{{v}_{\text{f}}-{v}_{0}}{{t}_{\text{f}}-{t}_{0}}$ .

## Solution

First, identify the knowns: ${v}_{0}=0,{v}_{\text{f}}=-15.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ (the negative sign indicates direction toward the west), Δ t = 1.80 s.

Second, find the change in velocity. Since the horse is going from zero to –15.0 m/s, its change in velocity equals its final velocity:

$\text{Δ}v={v}_{\text{f}}-{v}_{0}={v}_{\text{f}}=-15.0\phantom{\rule{0.2em}{0ex}}\text{m/s}.$

Last, substitute the known values ( $\text{Δ}v\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{Δ}t$ ) and solve for the unknown $\stackrel{\text{–}}{a}$ :

$\stackrel{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{-15.0\phantom{\rule{0.2em}{0ex}}\text{m/s}}{1.80\phantom{\rule{0.2em}{0ex}}\text{s}}=-8.33{\text{m/s}}^{2}.$

## Significance

The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s 2 due west means the horse increases its velocity by 8.33 m/s due west each second; that is, 8.33 meters per second per second, which we write as 8.33 m/s 2 . This is truly an average acceleration, because the ride is not smooth. We see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.

Check Your Understanding Protons in a linear accelerator are accelerated from rest to $2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ in 10 –4 s. What is the average acceleration of the protons?

Inserting the knowns, we have
$\stackrel{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m/s}-0}{{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{s}-0}=2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}{\text{m/s}}^{2}.$

## Instantaneous acceleration

Instantaneous acceleration a , or acceleration at a specific instant in time , is obtained using the same process discussed for instantaneous velocity. That is, we calculate the average velocity between two points in time separated by $\text{Δ}t$ and let $\text{Δ}t$ approach zero. The result is the derivative of the velocity function v ( t ), which is instantaneous acceleration    and is expressed mathematically as

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