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Calculating 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?

Two racehorses running toward the left.
(credit: Jon Sullivan, PD Photo.org)

Strategy

First we draw a sketch and assign a coordinate system to the problem. 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.

An acceleration vector arrow pointing west, in the negative x direction, labeled with a equals question mark. A velocity vector arrow also pointing toward the left, with initial velocity labeled as 0 and final velocity labeled as negative fifteen point 0 meters per second.

We can solve this problem by identifying Δ v and Δ t from the given information and then calculating the average acceleration directly from the equation a - = Δ v Δ t = v f v 0 t f t 0 .

Solution

1. Identify the knowns. v 0 = 0 , v f = 15 .0 m/s (the negative sign indicates direction toward the west), Δ t = 1 .80 s .

2. Find the change in velocity. Since the horse is going from zero to 15.0 m/s size 12{ - "15" "." 0`"m/s"} {} , its change in velocity equals its final velocity: Δ v = v f = 15 .0 m/s .

3. Plug in the known values ( Δ v and Δ t ) and solve for the unknown a - .

a - = Δ v Δ t = 15 .0 m/s 1 .80 s = 8 .33 m /s 2 .

Discussion

The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8 .33 m /s 2 due west means that 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 size 12{8 "." "33"`"m/s" rSup { size 8{2} } } {} . This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.

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Instantaneous acceleration

Instantaneous acceleration a , or the acceleration at a specific instant in time , is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed —that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. [link] shows graphs of instantaneous acceleration versus time for two very different motions. In [link] (a), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about 1 . 8 m /s 2 ). In [link] (b), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of + 3 . 0 m /s 2 and –2 . 0 m /s 2 , respectively.

Line graphs of instantaneous acceleration in meters per second per second versus time in seconds. The line on graph (a) shows slight variation above and below an average acceleration of about 1 point 8 meters per second per second. The line on graph (b) shows great variation over time, with instantaneous acceleration constant at 3 point 0 meters per second per second for 1 second, then dropping to negative 2 point 0 meters per second per second for the next 2 seconds, and then rising again, and so forth.
Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Here acceleration varies only slightly and is always in the same direction, since it is positive. The average over the interval is nearly the same as the acceleration at any given time. (b) Here the acceleration varies greatly, perhaps representing a package on a post office conveyor belt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as from 0 to 1.0 s) with constant or nearly constant acceleration in such a situation.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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