David Bauer Physics & Astronomy · UCLA

Physics 5A Chapter 5

Acceleration

Review: solving a kinematics problem

Before we introduce acceleration, let’s review the ideas of velocity and displacement by working through a kinematics problem, practicing the general problem-solving steps from the previous chapter.

Example: Two dogs meeting

Two dogs, Max and Luna, run toward each other. Luna runs east at $5.0\ \mathrm{m/s}$. Max starts $50\ \mathrm{m}$ east of Luna and runs west at $4.0\ \mathrm{m/s}$.

a. How long does it take the dogs to meet?

b. Where are they when they meet?

c. Sketch position vs. time graphs for the two dogs, and indicate the time and location the dogs meet on your graph.

Solution

We can use this problem to walk through the eight-step problem-solving process.

Step 1: Draw a picture. Sketch a horizontal $x$-axis pointing east. Luna is at the origin with a velocity arrow of magnitude $5.0\ \mathrm{m/s}$ pointing east (to the right), and Max is at $x = 50\ \mathrm{m}$ with a velocity arrow of magnitude $4.0\ \mathrm{m/s}$ pointing west (to the left).

A one-dimensional situation diagram. A horizontal arrow labeled x points to the right with east written below it. Luna, drawn as a filled circle, is at the origin x equals 0 with a green velocity arrow labeled v sub L equals 5.0 meters per second pointing to the right (east). Max, drawn as a filled circle, is 50 meters east of Luna with a green velocity arrow labeled v sub M equals 4.0 meters per second pointing to the left (west). A tick mark on the axis labels the position x equals 50 meters below Max.

Step 2: Choose a coordinate system. Take the $+x$ axis pointing east, with $x = 0$ at Luna’s starting position. With this choice, Luna moves in the $+x$ direction and Max moves in the $-x$ direction.

Step 3: Define symbols and list what you know. Let $d = 50\ \mathrm{m}$ be the initial distance between the dogs. Let $v_L = 5.0\ \mathrm{m/s}$ be Luna’s speed and $v_M = 4.0\ \mathrm{m/s}$ be Max’s speed.

Step 4: Identify the goal. We want the time $t_1$ when the dogs meet, and the position $x_1$ where they meet.

Step 5: Physical conditions. The dogs meet when they are at the same location, which means that their position functions are equal at time $t_1$: $x_L(t_1) = x_M(t_1)$.

Step 6: Write down relevant equations. Both dogs move in uniform motion, so their position functions are of the form

\[x(t) = x_0 + v_x t,\]

where $x_0$ is the initial position and $v_x$ is the $x$-component of the velocity. For Luna, $x_{L,0} = 0$ and $v_{L,x} = v_L$, so

\[x_L(t) = v_L t.\]

For Max, $x_{M,0} = d$ and $v_{M,x} = -v_M$ (negative because Max moves west), so

\[x_M(t) = d - v_M t.\]

Step 7: Solve. Setting the two position functions equal and solving for $t_1$,

\[\begin{aligned} v_L t_1 &= d - v_M t_1\\ \rightarrow v_L t_1 + v_M t_1 &= d\\ \rightarrow (v_L + v_M)\, t_1 &= d\\ \rightarrow t_1 &= \frac{d}{v_L + v_M} \end{aligned}\]

Plugging in the given numbers, we get

\[\begin{aligned} t_1 &= \frac{50\ \mathrm{m}}{5.0\ \mathrm{m/s} + 4.0\ \mathrm{m/s}}\\ &= \boxed{5.6\ \mathrm{s}.} \end{aligned}\]

To find the meeting position $x_1$, we can plug $t_1$ into either position function (since the dogs are at the same position at that time). Using Luna’s position,

\[\begin{aligned} x_1 &= v_L t_1\\ &= (5.0\ \mathrm{m/s})(5.6\ \mathrm{s})\\ &= \boxed{28\ \mathrm{m}.} \end{aligned}\]

Step 8: Check our answer.

  • Units: The time is in seconds and the position is in meters, which are the correct units for these quantities. ✓
  • Reasonable numbers: The dogs start 50 meters apart and run toward each other at (what we can assume are) realistic speeds for dogs, so it seems reasonable that they would meet after a few seconds. The meeting point is close to halfway between the dogs’ starting positions, and slightly closer to Max’s starting position, which makes sense since Luna is faster than Max. So the answers seem physically reasonable. ✓
  • Changing parameters: If the dogs started farther apart, we expect the time it would take them to reach each other to increase, and we see that $t_1$ is directly proportional to $d$. If the dogs ran faster, we expect the meeting time to decrease, and $t_1$ is inversely proportional to the sum of the speeds. So the answers have the correct dependence on the parameters. ✓

Part (c): position vs. time graphs. Both position functions are straight lines. Luna’s line starts at the origin with slope $+5.0\ \mathrm{m/s}$, and Max’s line starts at $50\ \mathrm{m}$ with slope $-4.0\ \mathrm{m/s}$. The two lines cross at the meeting point, $t_1 = 5.6\ \mathrm{s}$ and $x_1 = 28\ \mathrm{m}$, which you can mark with a dot and dashed lines connecting it to the axes.

A position versus time graph. The horizontal axis is labeled t in seconds with ticks from 1 to 8 seconds. The vertical axis is labeled x in meters with ticks at 10, 20, 30, 40, 50. Two straight lines are drawn. Luna's line labeled x sub L starts at the origin and rises linearly with slope 5 meters per second. Max's line labeled x sub M starts at x equals 50 meters on the vertical axis and decreases linearly with slope negative 4 meters per second. The two lines cross at a point near 5.6 seconds and 28 meters, which is marked with a dot. Dashed lines connect the crossing point to the axes.

This example demonstrates a common pattern in kinematics problems: solving for a time when some condition holds, and then plugging that time into a function of time to find other information. In this specific problem, we first solve for the time $t_1$ when the dogs are at the same position, so when $x_L(t) = x_M(t)$. Then we plug that time into the position equation $x_L(t)$ or $x_M(t)$ to find the position where they meet.

Instantaneous acceleration

Just like we defined the instantaneous velocity as the limit of the average velocity, we can define the instantaneous acceleration as the limit of the average acceleration. Recall that the average acceleration is defined as the change in velocity divided by the change in time:

\[\vec{a}_{\mathrm{av}} = \frac{\Delta \vec{v}}{\Delta t}\]

so, for example, the $x$-component of the average acceleration is

\[a_{\mathrm{av},x} = \frac{\Delta v_x}{\Delta t}.\]

If we take the limit as $\Delta t \to 0$, we get the $x$-component of the instantaneous acceleration:

\[\boxed{a_x = \frac{dv_x}{dt}}\]

Just as the instantaneous velocity is the slope of the position vs. time graph, the instantaneous acceleration is the slope of the velocity vs. time graph. A positive $a_x$ means $v_x$ is increasing:

Velocity v sub x versus time t graph showing a green straight line with positive slope, representing linearly increasing velocity under constant positive acceleration. No numerical tick labels.

A negative $a_x$ means $v_x$ is decreasing:

Velocity v sub x versus time t graph showing a green straight line with negative slope, starting positive and decreasing through zero. Represents linearly decreasing velocity under constant negative acceleration. No numerical tick labels.

Acceleration and position

Since the velocity is the derivative of the position,

\[v_x = \frac{dx}{dt},\]

the acceleration is the second derivative of the position:

\[\begin{aligned} a_x &= \frac{d}{dt}\left(\frac{dx}{dt}\right)\\ &= \frac{d^2 x}{dt^2} \end{aligned}\]

The second derivative tells us the curvature or concavity of the position vs. time graph:

  • If $a_x > 0$, the $x$ vs. $t$ graph is concave up.
  • If $a_x < 0$, the $x$ vs. $t$ graph is concave down.

A concave up position vs. time graph ($a_x > 0$) has a slope that increases with time (increasing velocity):

Position x versus time t graph showing a concave-up blue parabola with four green tangent line segments at evenly spaced times. Each successive tangent line is steeper than the previous, showing that the velocity increases over time.

A concave down position vs. time graph ($a_x < 0$) has a slope that decreases with time (decreasing velocity):

Position x versus time t graph showing a concave down blue parabola with five green tangent line segments. The tangent slopes decrease from steep positive at early times, through zero at the peak near t equals 2.5, to steep negative at later times.

Constant acceleration motion

Velocity in constant acceleration motion

Just like the displacement is the area under the $v_x(t)$ graph, the change in velocity is the area under the $a_x(t)$ graph. For constant acceleration, that area is a rectangle:

\[\Delta v_x = a_x \Delta t.\]

Acceleration a sub x versus time t graph showing a horizontal orange line at constant value a sub x. A shaded blue rectangle spanning a time interval Delta t under the line has area labeled Delta v sub x, representing the change in velocity. Brackets indicate a sub x on the y-axis and Delta t along the x-axis.

If a particle moving with constant acceleration $a_x$ starts with initial velocity $v_x(0) = v_{0x}$, then the velocity at time $t$ is

\[\boxed{v_x(t) = v_{0x} + a_x t.}\]

This is our constant acceleration velocity equation. Once we have the velocity equation, we can find the position from the area under the velocity vs. time graph.

Position in constant acceleration motion

For constant acceleration, the area under the $v_x(t)$ graph is a trapezoid. Decomposing it into a rectangle and triangle gives

\[\Delta x = v_{0x} t + \tfrac{1}{2} a_x t^2.\]

Velocity v sub x versus time t graph showing a green line starting at initial velocity v sub 0 and increasing linearly with slope a sub x. The trapezoidal area under the line is shaded blue and divided into a lower rectangle labeled v sub 0 times t and an upper triangle labeled one-half a sub x t squared. Brackets indicate v sub 0 on the y-axis and a sub x times t on the right side.

For a particle starting at initial position $x(0) = x_0$, the position at time $t$ is

\[\boxed{x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2.}\]

This is our constant acceleration position equation.

What we call the coordinates doesn’t matter

Don’t assume that because I am writing formulas for the $x$-component of position, velocity, and acceleration, that the motion has to be along the $x$-axis or in the horizontal direction! $x$ is just a placeholder for whatever coordinate we choose to describe the motion.

These same formulas apply to constant acceleration motion along any axis or in any direction. For example, along a $y$ coordinate axis, the formulas would be

\[\begin{aligned} y(t) &= y_0 + v_{0y} t + \frac{1}{2} a_y t^2,\\ v_y(t) &= v_{0y} + a_y t. \end{aligned}\]

Graphing constant acceleration motion

Since the position of an object in constant acceleration motion is a quadratic function of time, the position vs. time graph is a parabola. For $a_x > 0$, we get a concave up parabola:

Position x versus time t graph showing a concave-up blue parabola.

For $a_x < 0$, we get a concave down parabola:

Position x versus time t graph showing a concave down blue parabola with five green tangent line segments. The tangent slopes decrease from steep positive at early times, through zero at the peak near t equals 2.5, to steep negative at later times.

Example: Sketching $x(t)$ from a piecewise velocity graph

The plot below shows a velocity vs. time graph for an object that starts at $x = 0$ at $t = 0$. Sketch a position vs. time graph for the motion of the object from $t = 0$ to $t = 6$ seconds.

Velocity v sub x versus time t graph showing a piecewise linear function. The graph starts at v sub x equals 4 m/s at t equals 0, decreases linearly to v sub x equals negative 2 m/s at t equals 3 seconds, and then remains constant at v sub x equals negative 2 m/s until t equals 6 seconds. The graph has a green line color and a light grid in the background.

Solution

To sketch $x(t)$, we need the shape of the curve on each segment and the key times where the shape changes. The velocity graph has two segments:

  • From $t = 0$ to $t = 3\ \mathrm{s}$, $v_x$ is a straight line decreasing from $4\ \mathrm{m/s}$ to $-2\ \mathrm{m/s}$. The slope is constant, so this is constant-acceleration motion with

    \[a_x = \frac{-2\ \mathrm{m/s} - 4\ \mathrm{m/s}}{3\ \mathrm{s} - 0\ \mathrm{s}} = -2.0\ \mathrm{m/s^2}.\]

    Because $a_x$ is constant and negative, $x(t)$ is a concave-down parabola on this segment.

  • From $t = 3\ \mathrm{s}$ to $t = 6\ \mathrm{s}$, $v_x$ is constant at $-2.0\ \mathrm{m/s}$, so $a_x = 0$ and $x(t)$ is a straight line with slope $-2.0\ \mathrm{m/s}$.

Key positions.

  • Turning point at $t = 2\ \mathrm{s}$: the velocity crosses zero here, so the parabola reaches its peak.
  • Crossover at $t = 3\ \mathrm{s}$: where the parabola joins the straight line.

Drawing the concave-down parabola from $(0,\ 0)$ up to the turning point and down to the crossover, then the straight line of slope $-2.0\ \mathrm{m/s}$, we get the following sketch for $x(t)$.

A position x in meters versus time t in seconds graph. The horizontal time axis has ticks from 1 to 6 seconds and the vertical x axis has ticks from minus 3 to 4 meters. A blue curve starts at the origin, rises as a concave-down parabola to a maximum at the turning point t equals 2 seconds, x equals 4 meters, then curves back down and joins at t equals 3 seconds, x equals 3 meters, to a straight line segment. The straight line continues with constant negative slope to t equals 6 seconds, x equals negative 3 meters. Dashed reference lines mark the turning point.

Turning points in constant acceleration motion

A particle with velocity and acceleration in opposite directions will eventually stop and change direction at a turning point. At a turning point, the velocity is instantaneously zero.

Velocity v sub x versus time t graph showing a green straight line with negative slope, starting positive and crossing zero at the turning point time t sub turn. A dashed vertical line marks t sub turn on the time axis. An arrow labeled turning point points to the zero crossing. The region before t sub turn is labeled slowing down and the region after is labeled speeding up.

Position x versus time t graph showing a blue concave-down parabola that reaches a maximum at the turning point time t sub turn. A dashed vertical line marks t sub turn. A dashed horizontal line marks the maximum position x sub max. An arrow labeled turning point points to the peak. The region before t sub turn is labeled slowing down and the region after is labeled speeding up.

One example of a turning point is at the peak of the motion of a ball thrown straight up in the air. At the peak, the velocity is zero, and the ball changes direction from moving upward to moving downward.

The “$v^2$ equation”

There is a third equation for constant acceleration motion that is often useful, especially when we don’t know the time. We can derive it by solving for $t$ in the velocity equation and plugging that into the position equation:

\[\begin{aligned} \Delta t &= \frac{v_{f,x} - v_{i,x}}{a_x},\\ x_f &= x_i + v_{i,x} \Delta t + \frac{1}{2} a_x (\Delta t)^2. \end{aligned}\]

After some algebra, we get

\[\boxed{v_{f,x}^2 = v_{i,x}^2 + 2 a_x \Delta x.}\]

where $\Delta x = x_f - x_i$ is the displacement. There isn’t a standard name for this equation, so I usually just refer to it as the “$v$ squared equation.”

Example: Toy car with constant leftward acceleration

A toy car rolls on a horizontal surface and is initially moving right at $12.0\ \mathrm{m/s}$. The car has a constant acceleration directed to the left with magnitude $3.00\ \mathrm{m/s^2}$.

a. When does the car return to its starting point (after $t = 0$)?

b. What is the maximum distance to the right of its starting point that the car reaches?

Solution

Take $+x$ to the right, with the origin at the car’s starting point. We are given the initial position, velocity, and acceleration components,

\[x_0 = 0, \qquad v_{0x} = +12.0\ \mathrm{m/s}, \qquad a_x = -3.00\ \mathrm{m/s^2}.\]

The acceleration is negative because it points in the $-x$ direction. Since the acceleration is constant, we can use the constant-acceleration kinematic equations:

\[\begin{aligned} v_x(t) &= v_{0x} + a_x t,\\ x(t) &= x_0 + v_{0x}\,t + \tfrac{1}{2} a_x t^{2}. \end{aligned}\]

(a) The car is back at its starting point when $x(t) = 0$. Setting the position function to zero,

\[v_{0x}\,t + \tfrac{1}{2} a_x t^{2} = 0.\]

Since the problem asks for the time after $t = 0$, we can divide the equation by $t$ to get

\[v_{0x} + \tfrac{1}{2} a_x t = 0,\]

and rearranging gives the return time

\[t_{\mathrm{ret}} = -\frac{2 v_{0x}}{a_x} = -\frac{2\,(12.0\ \mathrm{m/s})}{-3.00\ \mathrm{m/s^2}} = \boxed{8.00\ \mathrm{s}.}\]

(b) The car reaches its maximum position at the turning point, where its velocity is instantaneously zero. From the velocity equation,

\[\begin{aligned} 0 &= v_{0x} + a_x t_{\mathrm{max}}\\ \rightarrow t_{\mathrm{max}} &= -\frac{v_{0x}}{a_x} = -\frac{12.0\ \mathrm{m/s}}{-3.00\ \mathrm{m/s^2}} = 4.00\ \mathrm{s}. \end{aligned}\]

Plugging this back into the position function,

\[\begin{aligned} x_{\max} &= x(t_{\mathrm{max}}) = v_{0x}\,t_{\mathrm{max}} + \tfrac{1}{2} a_x (t_{\mathrm{max}})^{2}\\ &= (12.0\ \mathrm{m/s})(4.00\ \mathrm{s}) - \tfrac{1}{2}(3.00\ \mathrm{m/s^2})(4.00\ \mathrm{s})^{2}\\ &= \boxed{24.0\ \mathrm{m}.} \end{aligned}\]

Alternatively, we can skip computing $t_{\mathrm{max}}$ altogether by using the $v^2$ equation with final velocity $v_x = 0$ at the turning point:

\[\begin{aligned} 0 &= v_{0x}^{2} + 2\,a_x\,\Delta x_{\max}\\ \rightarrow \Delta x_{\max} &= -\frac{v_{0x}^{2}}{2\,a_x} = -\frac{(12.0\ \mathrm{m/s})^{2}}{2\,(-3.00\ \mathrm{m/s^2})} = 24.0\ \mathrm{m}.\ \checkmark \end{aligned}\]

Both methods agree. Notice also that $t_{\mathrm{ret}} = 2\,t_{\mathrm{max}}$: by the symmetry of constant-acceleration motion, the trip out to the turning point and back takes equal times.

Symmetry of constant acceleration motion

In the previous example, we found that the time to return to the starting point is $t_{\mathrm{ret}} = 8.00\ \mathrm{s}$. The time to reach the maximum distance is exactly half of that, $t_{\mathrm{max}} = 4.00\ \mathrm{s} = t_{\mathrm{ret}}/2$.

This is not a coincidence, but a consequence of the symmetry of constant acceleration motion. The motion from $t = 0$ to $t = t_{\mathrm{max}}$ is a “mirror image” of the motion from $t = t_{\mathrm{max}}$ to $t = t_{\mathrm{ret}}$.

In particular, when an object moves in constant acceleration motion with the acceleration opposite the initial velocity, the time to reach the turning point ($t_{\mathrm{max}}$) is always half of the time to return to the starting point ($t_{\mathrm{ret}}$):

\[\boxed{t_{\mathrm{max}} = \frac{t_{\mathrm{ret}}}{2}.}\]

Position x versus time t graph showing a blue concave-down parabola that starts and ends at the same position x equals zero. The parabola reaches its maximum position x sub max at time t sub max, where a dashed vertical line marks the axis of symmetry. A dashed horizontal line marks x sub max on the vertical axis. A red dot marks the peak. Two equal-length double-headed arrows below the time axis show that the duration from t equals zero to t sub max equals the duration from t sub max to t sub ret, illustrating that t sub max equals t sub ret divided by two.