Lesson 1.2 · Unit 1 · CED topic 1.2
Displacement, velocity and acceleration
Kinematics is the description of motion, with no questions yet about what causes it. Three quantities do all the work: how far the object moved, how fast its position is changing, and how fast that is changing. They are displacement, velocity and acceleration, and each one is defined as a rate of change of the one before it. Students who learn the definitions precisely find the rest of Unit 1 almost mechanical; students who rely on the everyday meaning of “accelerate” (which in ordinary speech means only “speed up”) lose points on nearly every exam. Physics gives the word a sharper meaning, and this lesson builds it.
Describe a change in an object’s position. Describe the average velocity and acceleration of an object, and describe its velocity and acceleration.
- 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.
- 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. Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs.
- 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.
The object model
A real car has wheels, a roof and passengers; a real ball spins. To describe where something is and how it moves from place to place, none of that matters. The object model replaces the real thing with a single point that carries its mass. Position then means the position of that point. This simplification is why we can talk about “the position of a car” at all. It stops working when the shape or the rotation matters, for example in a question about a spinning wheel (Unit 5), and then the object must be treated as a rigid system instead.
Displacement and average velocity
The displacement of an object during an interval is its final position minus its initial position, \[ \Delta x = x - x_0 . \] The average velocity over the interval is the displacement divided by the time that the displacement took: \[ v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x - x_0}{t - t_0}. \] Its unit is meters per second. Because \(\Delta x\) is a vector component, so is the average velocity: it has the sign of the displacement. Notice what the definition leaves out. It looks only at the beginning and the end of the interval, so two objects can have the same average velocity while moving in completely different ways in between.
Acceleration is a change in velocity
The average acceleration over an interval is the change in velocity divided by the time of the change: \[ a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t - t_0}. \] The unit is meters per second per second, \(\text{m/s}^2\). An acceleration of \(+3\ \text{m/s}^2\) means that every second the velocity increases by \(3\ \text{m/s}\) in the positive direction. The velocity starts at 0, then it is \(+3\), then \(+6\), then \(+9\ \text{m/s}\), and so on.
Because velocity is a vector, an object accelerates whenever the magnitude of its velocity, the direction of its velocity, or both, change. Three different situations qualify. A car pulling away from a light speeds up. A car braking for a stop sign slows down. And a car going around a curve at a steady speedometer reading is also accelerating, because the direction of its velocity keeps changing (Unit 2 returns to this). In one dimension the direction can only reverse, and a reversal is a change of velocity just as surely as a change in speed.
A motion diagram like Figure 1.2a makes this visible. Dots show where the object was at equal time intervals, and arrows above the dots show the velocity. When the spacing of the dots is uniform the velocity is constant and the acceleration is zero. When the spacing grows, the velocity arrows grow, and the acceleration points in the same direction as the motion. When the spacing shrinks, the acceleration points opposite to the motion. The acceleration arrow always points in the direction of the change in velocity.
Speeding up or slowing down: compare the signs
The sign of the acceleration alone does not tell you whether an object speeds up. What matters is whether the velocity and the acceleration point the same way. If they have the same sign, the speed increases; if they have opposite signs, the speed decreases. A ball thrown upward (taking up as positive) has \(v \gt 0\) and \(a \lt 0\) on the way up and slows down. On the way down it has \(v \lt 0\) and \(a \lt 0\) and speeds up. The acceleration did not change at all; the velocity did.
Two further facts break common mistakes. First, zero velocity does not mean zero acceleration. At the instant a thrown ball reaches the top of its flight its velocity is zero, but its velocity is changing, so its acceleration is not zero (it is \(-10\ \text{m/s}^2\), as lesson 1.5 shows). Second, zero acceleration does not mean zero velocity: a car cruising at constant velocity on a straight road has \(a = 0\) with a large \(v\).
Instantaneous values from small intervals
Average velocity describes a whole interval, but a speedometer reads the velocity right now. The instantaneous velocity is what the average velocity approaches as the time interval is made very small. Suppose a car’s position is \(x = t^2\) in meters, with \(t\) in seconds. To find its velocity at \(t = 3.0\ \text{s}\), compute the average velocity from 3.0 s to a slightly later time and watch the value settle:
| Interval | \(\Delta x\) | \(\Delta t\) | average velocity |
|---|---|---|---|
| 3.0 s to 4.0 s | 7.0 m | 1.0 s | 7.0 m/s |
| 3.0 s to 3.5 s | 3.25 m | 0.5 s | 6.5 m/s |
| 3.0 s to 3.1 s | 0.61 m | 0.1 s | 6.1 m/s |
| 3.0 s to 3.01 s | 0.0601 m | 0.01 s | 6.01 m/s |
The averages close in on \(6.0\ \text{m/s}\), the instantaneous velocity at 3.0 s. The same idea works for acceleration: the average acceleration over a very small interval is very nearly the instantaneous acceleration. Lesson 1.3 shows the geometric version of this idea, where the instantaneous velocity is the slope of a tangent line.
A cyclist is at \(x = 12\ \text{m}\) when the clock reads 2.0 s and at \(x = 42\ \text{m}\) when it reads 7.0 s. Find the displacement and average velocity.
Step 1. \(\Delta x = 42 - 12 = +30\ \text{m}\) and \(\Delta t = 7.0 - 2.0 = 5.0\ \text{s}\).
Step 2. \(v_{\text{avg}} = \dfrac{+30\ \text{m}}{5.0\ \text{s}} = +6.0\ \text{m/s}\), in the positive direction.
Check. The cyclist moved a little over 4 m each second for 5 s, so the answer is in the right range if the speed were steady; it need not have been steady, only the average is known.
A car moving at \(+25\ \text{m/s}\) brakes steadily and slows to \(+10\ \text{m/s}\) in 6.0 s. Find its average acceleration and say what the sign means.
Step 1, change in velocity. \(\Delta v = 10 - 25 = -15\ \text{m/s}\).
Step 2. \(a_{\text{avg}} = \dfrac{-15\ \text{m/s}}{6.0\ \text{s}} = -2.5\ \text{m/s}^2\).
Meaning. The velocity is positive and the acceleration is negative, opposite signs, so the car slows down by 2.5 m/s each second. The acceleration points opposite to the motion.
A ball moving at \(+8.0\ \text{m/s}\) (toward a wall) bounces back at \(-6.0\ \text{m/s}\). The contact lasts 0.020 s. Find the average acceleration during the contact.
Step 1. \(\Delta v = v - v_0 = (-6.0) - (+8.0) = -14.0\ \text{m/s}\). The change is larger than either speed because the direction reversed: the ball had to be brought to rest and then sent back.
Step 2. \(a_{\text{avg}} = \dfrac{-14.0\ \text{m/s}}{0.020\ \text{s}} = -700\ \text{m/s}^2\).
Meaning. The acceleration is large because the time is short, and it is negative, pointing away from the wall. This is why impacts feel so violent: the changes of velocity are ordinary, but they happen in thousandths of a second.
- Subtract in the right order. \(\Delta v = v - v_0\), final minus initial. A reversed subtraction flips the sign of the answer and the conclusion about slowing or speeding up.
- Compare signs of \(v\) and \(a\). Same sign means speeding up; opposite signs mean slowing down. Write this reasoning in words on free-response questions.
- Acceleration is not “speed.” Never describe a constant acceleration as a constant speed or the reverse.
- Units. Velocity is m/s, acceleration is \(\text{m/s}^2\). A numerical answer without units loses a point.
- Direction can change without speed changing. A question may ask whether an object is accelerating while its speedometer is steady; the answer is yes if it is turning.
- A train is at \(x = 150\ \text{m}\) at \(t = 10\ \text{s}\) and at \(x = 30\ \text{m}\) at \(t = 30\ \text{s}\). Find its displacement and average velocity.
Show answer
\(\Delta x = 30 - 150 = -120\ \text{m}\), \(\Delta t = 20\ \text{s}\), so \(v_{\text{avg}} = -120/20 = -6.0\ \text{m/s}\). The train is moving, on average, in the negative direction at 6.0 m/s.
- A sprinter goes from rest to \(+9.0\ \text{m/s}\) in 1.5 s. Find her average acceleration.
Show answer
\(a_{\text{avg}} = (9.0 - 0)/1.5 = +6.0\ \text{m/s}^2\).
- An elevator moving at \(-4.0\ \text{m/s}\) (downward) stops in 2.0 s. Find the average acceleration, and state whether the elevator was speeding up or slowing down.
Show answer
\(a = (0 - (-4.0))/2.0 = +2.0\ \text{m/s}^2\). The velocity was negative and the acceleration is positive, opposite signs, so the elevator was slowing down.
- A student claims, “At the top of its flight a tossed ball has zero velocity, so it must have zero acceleration.” Evaluate the claim.
Show answer
The claim is wrong. Velocity is zero only for an instant, but it is changing from positive to negative through that instant, so the acceleration is not zero. In fact gravity gives the ball the same acceleration, about \(10\ \text{m/s}^2\) downward, at every point of its flight.
- A puck moving at \(+3.0\ \text{m/s}\) strikes a board and rebounds at \(-2.0\ \text{m/s}\) in 0.010 s. Find the average acceleration.
Show answer
\(\Delta v = -2.0 - 3.0 = -5.0\ \text{m/s}\). \(a_{\text{avg}} = -5.0/0.010 = -500\ \text{m/s}^2\).
- A car travels around a circular track at a constant speed of 20 m/s. Is it accelerating? Explain using the definition of acceleration.
Show answer
Yes. Acceleration is a change in the magnitude or the direction of the velocity. The speed is constant, but the direction of the velocity changes continuously as the car goes around, so the velocity vector changes and the car is accelerating.