The Central Idea
Acceleration is the rate at which velocity changes.
Because velocity includes both magnitude and direction, acceleration can occur when:
- speed increases;
- speed decreases;
- direction changes;
- both speed and direction change.
Therefore, acceleration does not simply mean “going faster.”
What Is Acceleration?
Acceleration measures the change in velocity per unit time.
Over a time interval, average acceleration is
\[ \boxed{ a_{\text{avg}} = \frac{\Delta v}{\Delta t} }. \]Since
\[ \Delta v=v_f-v_i, \]we may write
\[ \boxed{ a_{\text{avg}} = \frac{v_f-v_i}{t_f-t_i} }. \]Acceleration Is a Rate of a Rate
Velocity is already a rate:
\[ v = \frac{\text{change in position}}{\text{change in time}}. \]Acceleration measures how that rate changes:
\[ a = \frac{\text{change in velocity}}{\text{change in time}}. \]Acceleration can therefore be thought of as a rate of a rate.
Units of Acceleration
Velocity has units of length per time.
Dividing velocity by another unit of time gives
\[ \frac{\text{length}/\text{time}}{\text{time}} = \frac{\text{length}}{\text{time}^2}. \]Common acceleration units include:
- meters per second squared, \(\text{m/s}^2\);
- feet per second squared, \(\text{ft/s}^2\);
- kilometers per hour squared, \(\text{km/h}^2\).
What Does \(\text{m/s}^2\) Mean?
An acceleration of
\[ 3\frac{\text{m}}{\text{s}^2} \]means that velocity changes by
\[ 3\frac{\text{m}}{\text{s}} \]during each second, provided the acceleration remains constant.
If the initial velocity is \(2\text{ m/s}\), then under constant acceleration \(3\text{ m/s}^2\):
\[ v(0)=2, \] \[ v(1)=5, \] \[ v(2)=8, \] \[ v(3)=11. \]Example: Average Acceleration
A car's velocity changes from
\[ 10\frac{\text{m}}{\text{s}} \]to
\[ 22\frac{\text{m}}{\text{s}} \]in 4 seconds.
Then
\[ a_{\text{avg}} = \frac{22-10}{4} \] \[ = \frac{12}{4} \] \[ \boxed{ a_{\text{avg}} = 3\frac{\text{m}}{\text{s}^2} }. \]Example: Negative Average Acceleration
Suppose velocity changes from
\[ 18\frac{\text{m}}{\text{s}} \]to
\[ 6\frac{\text{m}}{\text{s}} \]in 3 seconds.
Then
\[ a_{\text{avg}} = \frac{6-18}{3} \] \[ = \frac{-12}{3} \] \[ \boxed{ a_{\text{avg}} = -4\frac{\text{m}}{\text{s}^2} }. \]Negative Acceleration Does Not Always Mean Slowing Down
The sign of acceleration describes direction relative to the chosen coordinate system.
It does not by itself tell whether speed is increasing or decreasing.
To determine whether an object is speeding up or slowing down, compare the signs of velocity and acceleration.
Velocity and Acceleration with the Same Sign
In one-dimensional motion, when velocity and acceleration have the same sign, the magnitude of velocity tends to increase.
The object is speeding up.
Two common cases are:
\[ v>0,\qquad a>0 \]and
\[ v<0,\qquad a<0. \]Velocity and Acceleration with Opposite Signs
When velocity and acceleration have opposite signs, the magnitude of velocity tends to decrease.
The object is slowing down.
Two common cases are:
\[ v>0,\qquad a<0 \]and
\[ v<0,\qquad a>0. \]A Useful Sign Table
| Velocity | Acceleration | Motion Tendency |
|---|---|---|
| \(v>0\) | \(a>0\) | Speeding up in the positive direction |
| \(v>0\) | \(a<0\) | Slowing down while moving in the positive direction |
| \(v<0\) | \(a<0\) | Speeding up in the negative direction |
| \(v<0\) | \(a>0\) | Slowing down while moving in the negative direction |
Example: Negative Velocity and Negative Acceleration
Suppose
\[ v=-5\frac{\text{m}}{\text{s}} \]and
\[ a=-2\frac{\text{m}}{\text{s}^2}. \]The object is moving in the negative direction.
The acceleration also points in the negative direction, so the magnitude of the velocity increases.
The object is speeding up, even though the acceleration is negative.
Example: Negative Velocity and Positive Acceleration
Suppose
\[ v=-8\frac{\text{m}}{\text{s}} \]and
\[ a=3\frac{\text{m}}{\text{s}^2}. \]The velocity points in the negative direction while the acceleration points in the positive direction.
The speed decreases until the velocity possibly reaches zero.
Average Acceleration over a Time Interval
If velocity is represented by a function \(v(t)\), then average acceleration from \(t_1\) to \(t_2\) is
\[ \boxed{ a_{\text{avg}} = \frac{v(t_2)-v(t_1)}{t_2-t_1} }. \]This is the slope of a secant line on a velocity-time graph.
Example: Average Acceleration from \(v(t)\)
Let
\[ v(t)=2t^2+1. \]Find the average acceleration from \(t=1\) to \(t=4\).
First,
\[ v(1)=3. \]Next,
\[ v(4)=33. \]Therefore,
\[ a_{\text{avg}} = \frac{33-3}{4-1} \] \[ = \frac{30}{3} \] \[ \boxed{a_{\text{avg}}=10}. \]Instantaneous Acceleration
Average acceleration describes velocity change over an interval.
Instantaneous acceleration describes how rapidly velocity is changing at one particular instant.
It is defined by the limit
\[ a(t) = \lim_{\Delta t\to0} \frac{v(t+\Delta t)-v(t)}{\Delta t}. \]Acceleration as a Derivative
If velocity is differentiable, then
\[ \boxed{ a(t)=v'(t)=\frac{dv}{dt} }. \]Since velocity is the derivative of position,
\[ v(t)=x'(t), \]acceleration is the second derivative of position:
\[ \boxed{ a(t)=x''(t)=\frac{d^2x}{dt^2} }. \]The Position-Velocity-Acceleration Chain
For one-dimensional motion,
\[ x(t) \xrightarrow{\text{differentiate}} v(t) \xrightarrow{\text{differentiate}} a(t). \]Symbolically,
\[ \boxed{ v(t)=x'(t), \qquad a(t)=v'(t)=x''(t) }. \]This chain is one of the most important structural relationships in kinematics.
Example: Acceleration from Position
Let
\[ x(t)=t^3-4t^2+2t+7. \]Differentiate once:
\[ v(t)=3t^2-8t+2. \]Differentiate again:
\[ a(t)=6t-8. \]Therefore,
\[ \boxed{ v(t)=3t^2-8t+2, \qquad a(t)=6t-8 }. \]Example: Instantaneous Acceleration
Continue with
\[ a(t)=6t-8. \]At \(t=3\),
\[ a(3)=18-8=10. \]Therefore,
\[ \boxed{a(3)=10}. \]Acceleration on a Velocity-Time Graph
On a velocity-time graph, acceleration is represented by slope.
Average acceleration is a secant slope:
\[ \frac{\Delta v}{\Delta t}. \]Instantaneous acceleration is the tangent slope at a particular time.
Positive Slope on a Velocity-Time Graph
If a velocity-time graph rises as time increases, then velocity is increasing.
Therefore,
\[ a>0. \]Whether the object is speeding up or slowing down still depends on the sign of velocity.
Negative Slope on a Velocity-Time Graph
If a velocity-time graph falls as time increases, then velocity is decreasing.
Therefore,
\[ a<0. \]Again, negative acceleration does not automatically mean decreasing speed.
Zero Slope on a Velocity-Time Graph
A horizontal velocity-time graph means velocity is constant.
Therefore,
\[ \boxed{a=0}. \]The object may still be moving. Zero acceleration means the velocity is not changing.
Acceleration on a Position-Time Graph
Acceleration is not read directly as the slope of a position-time graph.
The slope of a position-time graph is velocity.
Acceleration describes how that slope changes.
Thus acceleration is connected to the graph's concavity.
Concave Up Position Graphs
If a position-time graph is concave up, its slopes increase as time increases.
Therefore, acceleration is positive:
\[ a>0. \]Concave Down Position Graphs
If a position-time graph is concave down, its slopes decrease as time increases.
Therefore, acceleration is negative:
\[ a<0. \]Constant Acceleration
Constant acceleration means
\[ a(t)=a \]for a fixed value \(a\).
Under constant acceleration, velocity changes linearly with time.
This produces one of the standard kinematic equations:
\[ \boxed{ v_f=v_i+a\Delta t }. \]Deriving the Constant-Acceleration Velocity Formula
Begin with the definition of average acceleration:
\[ a = \frac{v_f-v_i}{\Delta t}. \]Multiply by \(\Delta t\):
\[ a\Delta t=v_f-v_i. \]Add \(v_i\):
\[ \boxed{ v_f=v_i+a\Delta t }. \]Example: Final Velocity under Constant Acceleration
An object begins with
\[ v_i=5\frac{\text{m}}{\text{s}} \]and accelerates at
\[ a=2\frac{\text{m}}{\text{s}^2} \]for
\[ \Delta t=6\text{ s}. \]Then
\[ v_f=v_i+a\Delta t \] \[ =5+2(6) \] \[ \boxed{ v_f=17\frac{\text{m}}{\text{s}} }. \]Example: Constant Negative Acceleration
Suppose
\[ v_i=24\frac{\text{m}}{\text{s}}, \] \[ a=-3\frac{\text{m}}{\text{s}^2}, \]and
\[ \Delta t=5\text{ s}. \]Then
\[ v_f=24+(-3)(5) \] \[ =9. \]Therefore,
\[ \boxed{ v_f=9\frac{\text{m}}{\text{s}} }. \]Velocity as a Linear Function under Constant Acceleration
If time is measured from \(t=0\), then
\[ \boxed{ v(t)=v_0+at }. \]This is a linear function of time.
On a velocity-time graph:
- the vertical intercept is \(v_0\);
- the slope is \(a\).
Average Velocity under Constant Acceleration
If acceleration is constant, velocity changes linearly.
Therefore, the average velocity over the interval is the arithmetic mean of the initial and final velocities:
\[ \boxed{ v_{\text{avg}} = \frac{v_i+v_f}{2} }. \]This shortcut depends on constant acceleration.
Example: Average Velocity under Constant Acceleration
Suppose velocity changes uniformly from
\[ 4\frac{\text{m}}{\text{s}} \]to
\[ 16\frac{\text{m}}{\text{s}}. \]Then
\[ v_{\text{avg}} = \frac{4+16}{2} \] \[ = 10\frac{\text{m}}{\text{s}}. \]Therefore,
\[ \boxed{ v_{\text{avg}} = 10\frac{\text{m}}{\text{s}} }. \]Displacement under Constant Acceleration
Since
\[ \Delta x = v_{\text{avg}}\Delta t, \]and
\[ v_{\text{avg}} = \frac{v_i+v_f}{2}, \]we obtain
\[ \boxed{ \Delta x = \frac{v_i+v_f}{2}\Delta t }. \]The Time-Based Displacement Formula
Substitute
\[ v_f=v_i+a\Delta t \]into
\[ \Delta x = \frac{v_i+v_f}{2}\Delta t. \]Then
\[ \Delta x = \frac{v_i+(v_i+a\Delta t)}{2}\Delta t. \]Simplifying,
\[ \boxed{ \Delta x = v_i\Delta t + \frac12 a(\Delta t)^2 }. \]Position under Constant Acceleration
Since
\[ \Delta x=x_f-x_i, \]the displacement formula can be written
\[ x_f-x_i = v_i\Delta t + \frac12 a(\Delta t)^2. \]Therefore,
\[ \boxed{ x_f = x_i + v_i\Delta t + \frac12 a(\Delta t)^2 }. \]Example: Displacement under Constant Acceleration
Suppose
\[ v_i=3\frac{\text{m}}{\text{s}}, \] \[ a=4\frac{\text{m}}{\text{s}^2}, \]and
\[ t=5\text{ s}. \]Then
\[ \Delta x = 3(5) + \frac12(4)(5^2) \] \[ = 15+50 \] \[ \boxed{ \Delta x=65\text{ m} }. \]A Constant-Acceleration Formula Without Time
Another useful relation is
\[ \boxed{ v_f^2 = v_i^2 + 2a\Delta x }. \]This equation is especially useful when time is unknown or unnecessary.
Like the other standard kinematic equations here, it assumes constant acceleration.
Deriving the Time-Free Formula
Start with
\[ v_{\text{avg}} = \frac{v_i+v_f}{2} \]and
\[ \Delta x = v_{\text{avg}}\Delta t. \]Also,
\[ \Delta t = \frac{v_f-v_i}{a}. \]Substitute:
\[ \Delta x = \frac{v_i+v_f}{2} \cdot \frac{v_f-v_i}{a}. \]Using the difference of squares,
\[ 2a\Delta x = v_f^2-v_i^2. \]Therefore,
\[ \boxed{ v_f^2=v_i^2+2a\Delta x }. \]Example: Solving without Time
An object begins at
\[ v_i=4\frac{\text{m}}{\text{s}} \]and accelerates at
\[ a=3\frac{\text{m}}{\text{s}^2} \]over a displacement of
\[ 10\text{ m}. \]Then
\[ v_f^2 = 4^2+2(3)(10) \] \[ = 16+60 \] \[ = 76. \]Thus
\[ \boxed{ v_f=\sqrt{76}=2\sqrt{19}\frac{\text{m}}{\text{s}} } \]for the positive-direction solution appropriate to this motion.
The Standard Constant-Acceleration Equations
For one-dimensional motion with constant acceleration,
\[ \boxed{ v_f=v_i+a\Delta t } \] \[ \boxed{ \Delta x = \frac{v_i+v_f}{2}\Delta t } \] \[ \boxed{ \Delta x = v_i\Delta t + \frac12 a(\Delta t)^2 } \] \[ \boxed{ v_f^2 = v_i^2+2a\Delta x }. \]Each equation relates a different subset of the same motion quantities.
Choosing a Constant-Acceleration Equation
A practical strategy is to list the known and unknown quantities:
- \(v_i\): initial velocity;
- \(v_f\): final velocity;
- \(a\): acceleration;
- \(\Delta t\): elapsed time;
- \(\Delta x\): displacement.
Then choose an equation containing the desired unknown and as many known quantities as possible.
Constant Acceleration Does Not Mean Constant Velocity
If
\[ a\ne0 \]and acceleration is constant, then velocity changes steadily with time.
Constant velocity occurs only when
\[ \boxed{a=0}. \]Zero Velocity Does Not Imply Zero Acceleration
An object can have zero velocity at one instant while acceleration remains nonzero.
For example, an object thrown vertically upward has
\[ v=0 \]at the highest point, but gravitational acceleration still acts.
Zero Acceleration Does Not Imply Zero Velocity
If
\[ a=0, \]velocity is constant.
That constant may be zero, but it need not be.
An object can move steadily at
\[ 20\frac{\text{m}}{\text{s}} \]while having zero acceleration.
Acceleration Can Change Direction
Acceleration itself can vary with time.
For example,
\[ a(t)=4t-8 \]is negative for \(t<2\), zero at \(t=2\), and positive for \(t>2\).
Therefore, constant-acceleration formulas should not be applied blindly to motion with variable acceleration.
Variable Acceleration
When acceleration varies with time, we use the function
\[ a=a(t). \]Velocity may then be recovered by accumulation:
\[ v(t) = v(t_0) + \int_{t_0}^{t}a(s)\,ds. \]Position can then be recovered by accumulating velocity.
Area under an Acceleration-Time Graph
Because
\[ a(t)=\frac{dv}{dt}, \]the signed area under an acceleration-time graph gives the change in velocity:
\[ \boxed{ \Delta v = \int_{t_1}^{t_2}a(t)\,dt }. \]Area under a Velocity-Time Graph
Similarly,
\[ v(t)=\frac{dx}{dt}, \]so the signed area under a velocity-time graph gives displacement:
\[ \boxed{ \Delta x = \int_{t_1}^{t_2}v(t)\,dt }. \]Thus slopes move downward through the derivative chain, while signed areas move upward through the accumulation chain.
Derivative and Accumulation Structure
Differentiation gives
\[ x(t) \rightarrow v(t) \rightarrow a(t). \]Accumulation works in the reverse direction:
\[ a(t) \rightarrow \Delta v \rightarrow \Delta x. \]This reciprocal structure is one reason kinematics is a natural setting for learning calculus.
Acceleration in Two Dimensions
In two dimensions, position is a vector:
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle. \]Velocity is
\[ \mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t),y'(t)\rangle. \]Acceleration is
\[ \boxed{ \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) = \langle x''(t),y''(t)\rangle }. \]Example: Vector Acceleration
Let
\[ \mathbf{r}(t) = \langle t^2,\,t^3-2t\rangle. \]Velocity is
\[ \mathbf{v}(t) = \langle 2t,\,3t^2-2\rangle. \]Acceleration is
\[ \boxed{ \mathbf{a}(t) = \langle 2,\,6t\rangle }. \]Three-Dimensional Acceleration
If
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]then
\[ \boxed{ \mathbf{a}(t) = \langle x''(t),y''(t),z''(t)\rangle }. \]Each component measures acceleration along one coordinate direction.
Acceleration Can Change Direction without Changing Speed
Because velocity is a vector, changing its direction is a change in velocity even if its magnitude remains constant.
Therefore, an object moving at constant speed along a curved path can still accelerate.
Uniform circular motion is the standard example.
Acceleration and Curved Motion
For curved motion, acceleration can be separated conceptually into effects that change:
- the magnitude of velocity;
- the direction of velocity.
Later in this section, Tangential and Normal Components will formalize this decomposition.
Uniform Circular Motion Preview
In uniform circular motion, speed remains constant while velocity continually changes direction.
The resulting acceleration points toward the center of the circle.
Its magnitude is
\[ a_c=\frac{v^2}{r}. \]This will be developed more fully in the later lesson on Uniform Circular Motion.
Gravity as an Acceleration
Near Earth's surface and neglecting air resistance, freely falling objects experience approximately constant downward acceleration.
If upward is chosen as positive, we often write
\[ a=-g, \]where
\[ g\approx9.8\frac{\text{m}}{\text{s}^2}. \]The negative sign represents the chosen downward direction.
Acceleration Is Coordinate-Dependent in Sign
Suppose gravity points downward.
If upward is positive, then
\[ a=-g. \]If downward is positive instead, then
\[ a=+g. \]The physical acceleration is unchanged; only its coordinate description changes.
Stopping under Constant Acceleration
If an object moves with initial velocity \(v_i\) and constant acceleration opposite the motion, the stopping time can be found by setting
\[ v_f=0. \]From
\[ v_f=v_i+a\Delta t, \]we obtain
\[ 0=v_i+a\Delta t. \]Therefore,
\[ \boxed{ \Delta t=-\frac{v_i}{a} }. \]Example: Stopping Time
A car travels at
\[ 20\frac{\text{m}}{\text{s}} \]and experiences constant acceleration
\[ -5\frac{\text{m}}{\text{s}^2}. \]Set final velocity equal to zero:
\[ 0=20-5t. \]Therefore,
\[ \boxed{t=4\text{ s}}. \]Stopping Distance
Once the stopping time is known, stopping displacement can be calculated using
\[ \Delta x = v_i t+\frac12 at^2. \]For the previous example,
\[ \Delta x = 20(4) + \frac12(-5)(4^2). \]Thus
\[ \Delta x = 80-40 \] \[ \boxed{\Delta x=40\text{ m}}. \]Common Error: Defining Acceleration as Speeding Up
Acceleration is not synonymous with increasing speed.
Acceleration means changing velocity.
That change may involve:
- speeding up;
- slowing down;
- turning;
- some combination of these.
Common Error: Treating Negative Acceleration as Slowing Down
Negative acceleration means acceleration points in the negative coordinate direction.
If velocity is also negative, the object can be speeding up.
Always compare the signs of \(v\) and \(a\).
Common Error: Confusing Velocity and Acceleration
Velocity measures the rate of change of position:
\[ v=x'. \]Acceleration measures the rate of change of velocity:
\[ a=v'=x''. \]The quantities are related, but they are not interchangeable.
Common Error: Assuming \(v=0\) Means \(a=0\)
An object can have zero velocity at an instant while acceleration remains nonzero.
A turning point is a common example.
Common Error: Assuming \(a=0\) Means the Object Is Stopped
Zero acceleration means velocity is not changing.
The velocity may be a nonzero constant.
Thus an object can travel steadily while
\[ a=0. \]Common Error: Using Constant-Acceleration Formulas for Variable Acceleration
Equations such as
\[ v_f=v_i+a\Delta t \]and
\[ \Delta x=v_i\Delta t+\frac12a(\Delta t)^2 \]assume that \(a\) is constant over the interval.
If \(a=a(t)\) varies, calculus or another appropriate method may be needed.
Common Error: Forgetting to Square Time
In
\[ \Delta x=v_i t+\frac12at^2, \]the time in the acceleration term is squared.
Replacing \(t^2\) with \(t\) changes both the numerical result and the units.
Common Error: Ignoring Units
Consider
\[ a=4\frac{\text{m}}{\text{s}^2} \]acting for
\[ 3\text{ s}. \]The velocity change is
\[ a\Delta t = 4\frac{\text{m}}{\text{s}^2}(3\text{ s}) \] \[ = 12\frac{\text{m}}{\text{s}}. \]The units confirm that acceleration multiplied by time produces a change in velocity.
Common Error: Losing the Direction Sign
If positive \(x\) points right and an acceleration points left, the acceleration must be represented with a negative sign.
Sign conventions should be chosen before substituting numbers into the equations.
A Reliable Procedure for Average Acceleration
- Choose or identify the positive direction.
- Record the initial velocity \(v_i\).
- Record the final velocity \(v_f\).
- Compute \(\Delta v=v_f-v_i\).
- Compute \(\Delta t=t_f-t_i\).
- Divide \(\Delta v\) by \(\Delta t\).
- Attach units of length per time squared.
- Interpret the sign in the chosen coordinate system.
A Reliable Procedure from a Position Function
- Start with \(x(t)\).
- Differentiate once to find \(v(t)\).
- Differentiate again to find \(a(t)\).
- Evaluate \(v(t)\) and \(a(t)\) at the requested time.
- Compare their signs if asked whether the object is speeding up or slowing down.
- Solve \(v(t)=0\) when possible turning times are needed.
A Reliable Procedure for Constant-Acceleration Problems
- Choose a positive direction.
- List \(v_i\), \(v_f\), \(a\), \(\Delta t\), and \(\Delta x\).
- Mark the known values and the desired unknown.
- Choose a constant-acceleration equation containing the needed quantities.
- Substitute values with signs and units.
- Solve algebraically.
- Check whether the sign and magnitude make physical and mathematical sense.
Example: Full Constant-Acceleration Workflow
A vehicle begins at
\[ v_i=8\frac{\text{m}}{\text{s}} \]and accelerates uniformly at
\[ a=2.5\frac{\text{m}}{\text{s}^2} \]for
\[ t=6\text{ s}. \]Final velocity:
\[ v_f=8+2.5(6) \] \[ =23\frac{\text{m}}{\text{s}}. \]Displacement:
\[ \Delta x = 8(6) + \frac12(2.5)(6^2) \] \[ = 48+45 \] \[ =93\text{ m}. \]Therefore,
\[ \boxed{ v_f=23\frac{\text{m}}{\text{s}}, \qquad \Delta x=93\text{ m} }. \]Example: Full Function Workflow
Let
\[ x(t)=2t^3-9t^2+12t+1. \]Velocity:
\[ v(t)=6t^2-18t+12. \]Factor:
\[ v(t)=6(t^2-3t+2) \] \[ =6(t-1)(t-2). \]Acceleration:
\[ a(t)=12t-18. \]At \(t=1.5\),
\[ a(1.5)=18-18=0. \]The object's acceleration is momentarily zero there, even though its velocity need not be zero.
Practice 1: Average Acceleration
A vehicle's velocity changes from \(6\text{ m/s}\) to \(18\text{ m/s}\) in 4 seconds. Find the average acceleration.
Practice 2: Negative Acceleration
A car's velocity changes from \(25\text{ m/s}\) to \(10\text{ m/s}\) in 5 seconds. Find the average acceleration.
Practice 3: Interpret the Signs
An object has
\[ v=-7\frac{\text{m}}{\text{s}} \]and
\[ a=-2\frac{\text{m}}{\text{s}^2}. \]Is the object speeding up or slowing down?
Practice 4: Acceleration from Velocity
Let
\[ v(t)=4t^2-3t+2. \]Find \(a(t)\) and \(a(2)\).
Practice 5: Velocity and Acceleration from Position
Let
\[ x(t)=t^3-3t^2+5t. \]Find \(v(t)\) and \(a(t)\).
Practice 6: Final Velocity
An object begins at \(4\text{ m/s}\) and accelerates constantly at \(3\text{ m/s}^2\) for 5 seconds. Find the final velocity.
Practice 7: Constant-Acceleration Displacement
An object has
\[ v_i=2\frac{\text{m}}{\text{s}}, \qquad a=4\frac{\text{m}}{\text{s}^2}, \qquad t=3\text{ s}. \]Find the displacement.
Practice 8: Solve without Time
An object has
\[ v_i=5\frac{\text{m}}{\text{s}}, \]constant acceleration
\[ a=2\frac{\text{m}}{\text{s}^2}, \]and displacement
\[ \Delta x=12\text{ m}. \]Find \(v_f\).
Practice 9: Vector Acceleration
Let
\[ \mathbf{r}(t) = \langle 3t^2,\;t^3-4t\rangle. \]Find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\).
Practice 10: Identify the Error
A student says:
“An object has velocity \(-10\text{ m/s}\) and acceleration \(-3\text{ m/s}^2\), so it must be slowing down because the acceleration is negative.”
Explain the error.
Practice Answers
Practice 1
\[ a_{\text{avg}} = \frac{18-6}{4} \] \[ = \frac{12}{4} \] \[ \boxed{ a_{\text{avg}} = 3\frac{\text{m}}{\text{s}^2} }. \]Practice 2
\[ a_{\text{avg}} = \frac{10-25}{5} \] \[ = \frac{-15}{5} \] \[ \boxed{ a_{\text{avg}} = -3\frac{\text{m}}{\text{s}^2} }. \]Practice 3
Velocity and acceleration are both negative, so they point in the same direction.
The magnitude of velocity is increasing.
Therefore, the object is speeding up.
Practice 4
\[ v(t)=4t^2-3t+2. \] \[ a(t)=v'(t)=8t-3. \]At \(t=2\),
\[ a(2)=16-3=13. \]Therefore,
\[ \boxed{ a(t)=8t-3, \qquad a(2)=13 }. \]Practice 5
\[ x(t)=t^3-3t^2+5t. \] \[ v(t)=3t^2-6t+5. \] \[ a(t)=6t-6. \]Therefore,
\[ \boxed{ v(t)=3t^2-6t+5, \qquad a(t)=6t-6 }. \]Practice 6
\[ v_f=v_i+at \] \[ =4+3(5) \] \[ \boxed{ v_f=19\frac{\text{m}}{\text{s}} }. \]Practice 7
\[ \Delta x = v_i t+\frac12at^2 \] \[ = 2(3)+\frac12(4)(3^2) \] \[ = 6+18 \] \[ \boxed{ \Delta x=24\text{ m} }. \]Practice 8
\[ v_f^2 = v_i^2+2a\Delta x \] \[ = 5^2+2(2)(12) \] \[ = 25+48 \] \[ = 73. \]Therefore,
\[ \boxed{ v_f=\sqrt{73}\frac{\text{m}}{\text{s}} } \]for the positive-direction solution.
Practice 9
\[ \mathbf{r}(t) = \langle 3t^2,\;t^3-4t\rangle. \] \[ \mathbf{v}(t) = \langle 6t,\;3t^2-4\rangle. \] \[ \mathbf{a}(t) = \langle 6,\;6t\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}(t) = \langle 6t,\;3t^2-4\rangle, \qquad \mathbf{a}(t) = \langle 6,\;6t\rangle }. \]Practice 10
The student is treating negative acceleration as though it automatically meant decreasing speed.
Here,
\[ v<0 \]and
\[ a<0. \]Velocity and acceleration have the same sign, so the magnitude of velocity is increasing.
The object is therefore speeding up in the negative direction.
What to Remember
Average acceleration is
\[ \boxed{ a_{\text{avg}} = \frac{\Delta v}{\Delta t} }. \]Instantaneous acceleration is
\[ \boxed{ a(t)=v'(t)=x''(t) }. \]Under constant acceleration,
\[ \boxed{ v_f=v_i+a\Delta t } \] \[ \boxed{ \Delta x = v_i\Delta t+\frac12a(\Delta t)^2 } \] \[ \boxed{ v_f^2=v_i^2+2a\Delta x }. \]The key interpretive rules are:
- acceleration measures change in velocity, not simply increase in speed;
- velocity and acceleration with the same sign usually mean speeding up;
- velocity and acceleration with opposite signs usually mean slowing down;
- zero velocity does not necessarily mean zero acceleration;
- zero acceleration does not necessarily mean zero velocity;
- changing direction is a form of acceleration even at constant speed.
Where We Go Next
Position, velocity, and acceleration give the core quantities of motion. The next step is to describe motion whose coordinates are written directly as functions of a shared parameter.
The next lesson, Parametric Motion, develops motion through equations such as \(x=x(t)\) and \(y=y(t)\), connecting paths, timing, velocity, and acceleration in multiple dimensions.