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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.

Questions & Answers

calculate the tension of the cable when a buoy with 0.5m and mass of 20kg
Iga Reply
what is displacement
Nyamza Reply
what is the meaning of physics
Alausa Reply
to study objects in motion and how they interact or take part in the natural phenomenon of the universe.
Phill
an object that has a small mass and an object has a large mase have the same momentum which has high kinetic energy
Faith Reply
The with smaller mass
Gift
how
Faith
Since you said they have the same momentum.. So meaning that there is more like an inverse proportionality in the quantities used to find the momentum. We are told that the the is a larger mass and a smaller mass., so we can conclude that the smaller mass had higher velocity as compared to other one
Gift
Mathamaticaly correct
megavado
Mathmaticaly correct :)
megavado
I have proven it by using my own values
Gift
Larger mass=4g Smaller mass=2g Momentum of both=8 Meaning V for L =2 and V for S=4 Now find there kinetic energies using the data presented
Gift
grateful soul...thanks alot
Faith
Welcome
Gift
2 stones are thrown vertically upward from the ground, one with 3 times the initial speed of the other. If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? If the slower stone reaches a maximum height of H, how high will the faster stone go
Julliene Reply
30s
Gift
is speed the same as velocity
Faith Reply
no
Nebil
in a question i ought to find the momentum but was given just mass and speed
Faith
just multiply mass and speed then you have the magnitude of momentem
Nebil
Yes
Gift
Consider speed to be velocity
Gift
it worked our . . thanks
Faith
Distinguish between semi conductor and extrinsic conductors
Okame Reply
Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
Aj Reply
I think you shorten am not sure
Uche
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
shorten the pendulum string because the difference in length affects the time of oscillation.if short , the time taken will be adjusted.but if long ,the time taken will be twice the previous cycle.
FADILAT
discuss under damped
Prince Reply
resistance of thermometer in relation to temperature
Ifeanyi Reply
how
Bernard
that resistance is not measured yet, it may be probably in the next generation of scientists
Paul
Is fundamental quantities under physical quantities?
Igwe Reply
please I didn't not understand the concept of the physical therapy
John Reply
physiotherapy - it's a practice of exercising for healthy living.
Paul
what chapter is this?
Anderson
this is not in this book, it's from other experiences.
Paul
am new in the group
Daniel
please I have probably with calculate please can you please and help me out
John Reply
Sure
Gift
What is Boyce law
Sly Reply
Boyles law states that the volume of a fixed amount of gas is inversely proportional to pressure acting on that given gas if the temperature remains constant which is: V<k/p or V=k(1/p)
FADILAT
how to convert meter per second to kilometers per hour
grace Reply
Divide with 3.6
Mateo
multiply by (km/1000m) x (3600 s/h) -> 3.6
Muhammad
Practice Key Terms 4

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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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