Analyse displacement, velocity and a complete straight-line journey

Given

\(\displaystyle x(t)=t^2-4t+3,\quad v(t)=2t-4\)

Find the velocity at one second, in metres per second

Visit every time and compare position, velocity and acceleration. The position value is not its rate

At two seconds, which statement holds?

Two derivatives describe three different quantities

Differentiate position once for velocity and velocity once for acceleration. In this example the velocity changes linearly and acceleration is constant

1. 1Differentiate the right quantity Velocity is the derivative of signed displacement with respect to time. Acceleration is the derivative of velocity. Their units are length per time and length per time squared

Given

\(\displaystyle v(t)=2t-4\)

Find the acceleration at three seconds, in metres per second squared

At one second, velocity is −2 m/s. What is the speed?

Given

\(\displaystyle v(t)=2t-4\)

Find speed at t = 3 seconds, in metres per second

Solve velocity zero on the time interval

Rest occurs when 2t − 4 = 0, so t = 2. This time lies inside 0 ≤ t ≤ 4. Velocity is negative before it and positive afterwards, so direction reverses

1. 1Use velocity for direction and rest Positive velocity means motion in the positive direction; negative velocity means motion in the negative direction. Rest means velocity is zero, even when displacement is nonzero
2. 2Check both sides of a rest time A zero of velocity is a direction reversal only when velocity changes sign. A repeated root can give momentary rest without reversal

Given

\(\displaystyle v(t)=2t-4,\quad 0\le t\le4\)

At what time is the object at rest?

Given

\(\displaystyle x(t)=t^2-4t+3,\quad t=2\)

Find displacement at that rest time, in metres

Does negative displacement always mean motion in the negative direction?

Add distances between event positions

The positions at times 0, 2 and 4 are 3, −1 and 3 metres. The first segment travels 4 m and the second travels another 4 m.

1. 1Partition the journey Find every velocity zero inside the time interval. Evaluate displacement there and at the endpoints, then sum the absolute changes. Splitting at a rest time without reversal is harmless

Given

\(\displaystyle x(t)=t^2-4t+3,\quad 0\le t\le4,\quad x(0)=3,\quad x(2)=-1,\quad x(4)=3\)

Find total distance over the four-second journey, in metres

A return journey has different average rates

The object ends where it started, so net displacement change is zero. It still travels 8 metres in 4 seconds

1. 1Keep the average numerators distinct Average velocity is net displacement change divided by elapsed time. Average speed is total distance divided by elapsed time. A return journey can have zero average velocity and positive average speed

Given

\(\displaystyle x(0)=3,\quad x(4)=3,\quad T=4\)

Find average velocity over the whole journey, in metres per second

Given

\(\displaystyle D=8,\quad T=4\)

Find average speed over the whole journey, in metres per second

Zero acceleration can accompany continuing motion

For x(t) = 5 − 3t, velocity is the nonzero constant −3. The object moves in the negative direction at speed 3; it never rests

1. 1Differentiate the right quantity Velocity is the derivative of signed displacement with respect to time. Acceleration is the derivative of velocity. Their units are length per time and length per time squared
2. 2Use velocity for direction and rest Positive velocity means motion in the positive direction; negative velocity means motion in the negative direction. Rest means velocity is zero, even when displacement is nonzero

Given

\(\displaystyle x(t)=5-3t\)

Find acceleration for this constant-velocity motion

A repeated velocity zero does not reverse direction

For the position and velocity rules shown, velocity is positive on both sides of t = 2. The object rests momentarily and continues in the same direction

1. 1Check both sides of a rest time A zero of velocity is a direction reversal only when velocity changes sign. A repeated root can give momentary rest without reversal

Given

\(\displaystyle x(t)=\frac{t^3}3-2t^2+4t+5,\quad t=2\)

Find displacement at this momentary rest time

Given

\(\displaystyle v(t)=(t-2)^2\)

At t = 2 in this repeated-root model, what happens?

Match position, velocity and acceleration graphs

For the first model, position is an upward parabola with minimum at t = 2. Velocity is an increasing straight line crossing zero at 2. Acceleration is the horizontal line at 2. To choose position from a velocity rule, differentiate each candidate and check its stated initial value

1. 1Match derivative features The slope of the position graph gives velocity; the slope of the velocity graph gives acceleration. Use exact candidate rules, zero locations, signs and initial values. Do not copy graph heights between different quantities

Which description of the acceleration graph matches velocity v(t) = 2t − 4?

Keep rational rest times exact

The displayed velocity factor gives the exact rest time shown. Use that value when calculating event positions and segment lengths

1. 1Partition the journey Find every velocity zero inside the time interval. Evaluate displacement there and at the endpoints, then sum the absolute changes. Splitting at a rest time without reversal is harmless

Given

\(\displaystyle v(t)=2t-3,\quad0\le t\le4\)

Find the exact rest time on this interval