CourseCourseQ BankQuestion BankDocsDocuments

1.2 Displacement, Velocity, and Acceleration

Syllabus
2024
Topic
1.2
Level
—

Learning objectives

1.2A—Describe a change in an object’s positionDescribe a change in an object’s position.• When using the object model, the size, shape, and internal configuration are ignored. The object may be treated as a single point with extensive properties such as mass and charge.• Displacement is the change in an object’s position. Relevant equation: TOPIC 1.2 Displacement, Velocity, and Acceleration1.2B—Describe the average velocity and acceleration of an objectDescribe the average velocity and acceleration of an object.• Averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time.• Average velocity is the displacement of an object divided by the interval of time in which that displacement occurs. Relevant equation: • Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs. Relevant equation:  AP Physics 1: Algebra-Based Course and Exam Description Kinematics UNIT 1• An object is accelerating if the magnitude and/or direction of the object’s velocity are changing.• Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration. AP Physics 1: Algebra-Based Course and Exam Description Kinematics UNIT 1

Displacement compares two positions

Model the object

In the object model, treat the object as a single point. Its size, shape and internal details are ignored so its position can be tracked on a chosen axis.

Compare the endpoints

Δx=x−x0\Delta x=x-x_0

Hypothetical example: choose right as positive. An object moves from x0=+2 mx_0=+2\,\mathrm{m} to x=−3 mx=-3\,\mathrm{m}. Then Δx=(−3 m)−(+2 m)=−5 m\Delta x=(-3\,\mathrm{m})-(+2\,\mathrm{m})=-5\,\mathrm{m}. The magnitude is 5 m and the negative sign means the displacement is to the left.

Do not confuse path and change

Displacement depends only on the initial and final positions; distance traveled depends on the whole path. Returning to the starting point gives zero displacement even when the distance traveled is not zero.

Average rates compare a change with a time interval

Match each rate to its change

Average quantity Change in the numerator Relationship SI unit
Velocity Displacement, Δx\Delta x vˉ=ΔxΔt\bar v=\dfrac{\Delta x}{\Delta t} m s−1\mathrm{m\,s^{-1}}
Acceleration Velocity, Δv=v−v0\Delta v=v-v_0 aˉ=ΔvΔt\bar a=\dfrac{\Delta v}{\Delta t} m s−2\mathrm{m\,s^{-2}}

Substitute with signs and units

vˉ=+12 m4 s=+3 m s−1\bar v=\frac{+12\,\mathrm{m}}{4\,\mathrm{s}}=+3\,\mathrm{m\,s^{-1}}

In the same hypothetical interval, suppose velocity changes from v0=+4 m s−1v_0=+4\,\mathrm{m\,s^{-1}} to v=−4 m s−1v=-4\,\mathrm{m\,s^{-1}}. Then Δv=−8 m s−1\Delta v=-8\,\mathrm{m\,s^{-1}} and aˉ=(−8 m s−1)/(4 s)=−2 m s−2\bar a=(-8\,\mathrm{m\,s^{-1}})/(4\,\mathrm{s})=-2\,\mathrm{m\,s^{-2}}. The negative sign means the average acceleration points in the negative direction.

Interpret acceleration carefully

Acceleration occurs when velocity changes in magnitude and/or direction. Negative acceleration does not automatically mean slowing down. As the time interval becomes very small, an average value approaches the corresponding instantaneous value.