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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] .
We can solve this problem by identifying $\text{\Delta}v\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{\Delta}t$ from the given information, and then calculating the average acceleration directly from the equation $\stackrel{\text{\u2013}}{a}=\frac{\text{\Delta}v}{\text{\Delta}t}=\frac{{v}_{\text{f}}-{v}_{0}}{{t}_{\text{f}}-{t}_{0}}$ .
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:
Last, substitute the known values ( $\text{\Delta}v\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{\Delta}t$ ) and solve for the unknown $\stackrel{\text{\u2013}}{a}$ :
Check Your Understanding Protons in a linear accelerator are accelerated from rest to $2.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ in 10 ^{–4} s. What is the average acceleration of the protons?
Inserting the knowns, we have
$\stackrel{\text{\u2013}}{a}=\frac{\text{\Delta}v}{\text{\Delta}t}=\frac{2.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m/s}-0}{{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{s}-0}=2.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{11}{\text{m/s}}^{2}.$
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 $\text{\Delta}t$ and let $\text{\Delta}t$ 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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