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$m\,dv/dt = F$ solved for forces that depend on time, velocity or position, with $a = v\,dv/dx$ and the work–energy theorem as its integral.
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By the end of this lesson you will be able to write Newton's second law as a differential equation and solve it for forces that depend on time, velocity or position.
From Physics 1 you can draw free-body diagrams and use $F_{\text{net}} = ma$ for constant forces, where the constant-acceleration equations do the rest. The last two lessons connected position, velocity and acceleration by derivatives and integrals. From calculus you know how to separate variables and integrate. This lesson treats Newton's second law as what it really is, a differential equation, and solves it when the force changes.
| Term | What it means |
|---|---|
| Net force | The vector sum of all forces on the object; the second law uses only this. |
| Differential equation | An equation relating a function to its derivatives, such as $m\,dv/dt = F$. |
| Separation of variables | Rearranging so each side contains only one variable, then integrating both. |
| Chain rule substitution | $a = v\,dv/dx$, used when the force depends on position. |
| Initial condition | A known starting value that fixes the constant of integration. |
| Impulse | $\int F\,dt$, the change in momentum produced by a force over time. |
| Terminal state | A condition, such as constant velocity, that a solution approaches as time grows. |
The second law, $F_{\text{net}} = ma$, is a statement about rates:
$$m\frac{dv}{dt} = F_{\text{net}}.$$
When the net force is constant this integrates to the familiar constant-acceleration equations. When it changes, the method depends on what the force depends on:
The third form, integrated, is $\tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 = \int F\,dx$: the work–energy theorem, derived from Newton's law. In each case the constant of integration comes from the starting state, and the answer can be checked by substituting it back into $m\,dv/dt = F$.
Another way: picture
Think of the net force as a hand on the throttle. If the hand follows a clock, you add up its pushes over time. If it responds to how fast you are going, as drag does, you ask how long each change of speed takes. If it depends on where you are, as a spring does, you ask how much each meter of travel changes your speed. Same law, three bookkeeping schemes.
Another way: steps
A force given as a function of time, such as a rocket's thrust schedule or a push that ramps up, is the simplest case. Dividing by the mass gives the acceleration as a function of time, and the previous lesson's integrations give $v(t)$ and $x(t)$. For $F = ct$ from rest, $v = ct^2/2m$ and $x = ct^3/6m$.
The integral $\int F\,dt$ is the impulse, so the change in velocity is the impulse over the mass. When only the final velocity is wanted, the impulse is often easiest found as an area under the force–time graph: a thrust that falls linearly from $F_0$ to zero over a time $T$ delivers $\tfrac{1}{2}F_0T$. The momentum lesson develops this; here it is just the first integration of the second law.
Air and water resist motion with forces that grow with speed. Then $m\,dv/dt = F(v)$ has the velocity on both sides, and it is solved by separating: move everything with $v$ to one side and $dt$ to the other, $\frac{m\,dv}{F(v)} = dt$, and integrate.
For a block sliding with linear drag, $F = -bv$, this gives $\frac{m}{b}\ln(v_0/v) = t$, so $v = v_0e^{-bt/m}$: the velocity decays exponentially. For an object falling with drag, the net force $mg - bv$ vanishes at the terminal velocity $mg/b$, and the velocity approaches it. Lesson 15 develops these fully; the point here is that the separation is always possible when the force depends on velocity alone.
Springs, gravity at a distance and electric forces depend on where the object is. Then $F(x)$ is known but $x(t)$ is not, so neither side can be integrated over time directly. The chain rule rescues it: $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$. The second law becomes $mv\,dv = F(x)\,dx$, with each side in one variable.
Integrating gives $\tfrac{1}{2}mv^2\big|_{v_0}^{v} = \int_{x_0}^{x}F\,dx$: the change in kinetic energy equals the work done. This is how the work–energy theorem follows from Newton's law, and why energy methods are the natural tool for forces of position. The next lesson builds on it with variable forces and work.
The equations $v = v_0 + at$ and $x = x_0 + v_0t + \tfrac{1}{2}at^2$ were derived by integrating a constant $a$. If the force changes, $a$ changes, and those integrations are wrong. The error is not small: for a force that rises linearly from zero, using the final acceleration throughout overestimates the distance by a factor of three; using the average acceleration is closer but still wrong for the position.
A quick test before any calculation: does anything in the problem make the net force change? A rope's tension that is adjusted, a spring that stretches, drag that grows with speed or a rocket that burns fuel all do. If so, write the differential equation. If not, the familiar equations are exact and quickest.
Real problems usually combine forces. A box dragged across a floor by a rope whose tension grows in time also feels constant kinetic friction: $m\,dv/dt = T(t) - \mu_kmg$. This is still a force of time, so integrate directly, but only once the box starts moving. While $T(t) < \mu_smg$, static friction holds it still, and the equation does not yet apply.
When forces of different kinds combine, such as a spring force $-kx$ with drag $-bv$, the equation $m\ddot{x} = -kx - b\dot{x}$ can no longer be separated in either form. It is a linear second-order equation, solved with exponentials and sinusoids in the oscillation lessons. Recognizing which kind of equation you have is the first and most important step.
Checking an answer. Differentiate your $v(t)$ and multiply by $m$; you must get the force. Set the variable force to a constant, and your result must reduce to a constant-acceleration formula. Units must balance: $\int F\,dt$ is N s, which is kg m/s. And the direction of change must match the force: a positive force never slows an object moving in the positive direction.
Many real forces, such as the drag on a baseball with its speed-squared law in two dimensions, give equations with no formula solution. Then the second law is integrated numerically. Euler's method takes a small time step $\Delta t$ and updates $v \leftarrow v + (F/m)\Delta t$ and $x \leftarrow x + v\Delta t$, over and over.
Smaller steps give better answers, and better methods, such as the Runge–Kutta family, reach good accuracy with larger steps. The engineering software that predicts a car's crash response or a spacecraft's trajectory does exactly this, stepping Newton's second law forward in time for thousands of parts at once. The differential equation is the model; the numerical method is how it is solved when calculus alone cannot.
The second law applies to whatever system you choose, and the choice decides which forces appear. For two blocks joined by a string and pulled across a floor, treat each block as its own system and the string's tension appears in both equations, equal and opposite. Treat the pair as one system and the tension is internal and drops out, leaving $(m_1 + m_2)\,dv/dt = F_{\text{external}}$.
When the external force depends on time, the combined equation is often the quickest route to the motion, and the separate equations then give the tension. When the forces on the parts differ in kind, as when one block feels drag and the other does not, the parts must be treated separately and the equations solved together. In every case, write the second law for a clearly stated system, list only the forces that act on that system from outside it, and then decide what the net force depends on. Most mistakes in these problems come from including an internal force, or leaving out an external one, rather than from the calculus.
The high-speed test track at Holloman Air Force Base in New Mexico is a rail almost $16$ km long on which rocket-powered sleds reach speeds above $2800$ m/s to test ejection seats, missile components and parachutes. A sled's thrust is not constant: solid rocket motors often start strong and taper as their propellant burns.
A thrust falling linearly from $F_0$ to zero in time $T$ gives the sled an impulse of $\tfrac{1}{2}F_0T$, so from rest $v = F_0T/2m$. A $1000$ kg sled whose thrust falls from $100$ kN over $4$ s leaves the burn at $200$ m/s, half what a steady $100$ kN would give. Engineers integrate the actual thrust curves, measured on static test stands, together with drag and rail friction, to predict each run's speed profile before it is fired.
Launched roller coasters, such as those at Cedar Point in Ohio and Six Flags parks, accelerate trains with linear induction motors or hydraulic cables instead of a lift hill. The force is programmed as a function of time: it rises quickly, holds, and tapers so riders are not jerked at the start or end.
Designers integrate that force profile, together with the growing air drag and wheel friction, to find the train's speed at the end of the launch, which must be just enough to clear the first tower. Too little and the train rolls back, which coasters are designed to allow safely; too much wastes energy and stresses the track. Every launch is a solution of $m\,dv/dt = F(t) - F_{\text{drag}}(v) - F_{\text{friction}}$, integrated numerically with the measured forces.
It is tempting to reach for $x = x_0 + v_0t + \tfrac{1}{2}at^2$ in every mechanics problem. Those equations were derived for constant acceleration and give wrong answers, often badly wrong, when the net force changes. A spring, drag, a rocket's thrust or a rope pulled harder and harder all change the force, and each needs the differential equation.
A second error is to integrate a force of position over time, $\int F(x)\,dt$, as though $x$ were known. Use $a = v\,dv/dx$ instead, which trades time for position.
A $2.0$ kg cart starts from rest and is pushed with $F = 12t$ (N, s). Write the second law.
$2.0\dfrac{dv}{dt} = 12t$
The force depends on time only.
Solve for the acceleration.
$\dfrac{dv}{dt} = 6t$
Divide by the mass.
Integrate for the velocity from rest.
$v = 3t^2$
The antiderivative of $6t$.
Integrate for the position from the origin.
$x = t^3$
The antiderivative of $3t^2$.
Evaluate after $2.0$ s.
$v = 12\ \text{m/s}, \quad x = 8.0\ \text{m}$
Substitute the time.
A $0.50$ kg puck slides at $6.0$ m/s on a surface giving drag $F = -0.25v$ (N). Write the second law.
$0.50\dfrac{dv}{dt} = -0.25v$
The force depends on velocity.
Separate the variables.
$\dfrac{dv}{v} = -0.50\,dt$
Divide by $0.50v$.
Integrate with limits.
$\ln\dfrac{v}{6.0} = -0.50t$
From $6.0$ m/s at $t = 0$.
Solve for the velocity.
$v = 6.0e^{-0.50t}\ \text{m/s}$
Exponentiate.
Find the speed after $2.0$ s.
$v = 6.0e^{-1} = 2.2\ \text{m/s}$
It never quite stops.
A $2.0$ kg block passes $x = 0$ at $2.0$ m/s, pushed by $F = 8x$ (N, m). Write the chain-rule form.
$2.0v\dfrac{dv}{dx} = 8x$
The force depends on position.
Separate the variables.
$v\,dv = 4x\,dx$
Divide by the mass.
Integrate with limits.
$\tfrac{1}{2}v^2 - \tfrac{1}{2}(2.0)^2 = 2x^2$
From $2.0$ m/s at the origin.
Solve for the speed squared.
$v^2 = 4.0 + 4x^2$
Multiply by two.
Find the speed at $x = 3.0$ m.
$v = \sqrt{4.0 + 36} = 6.3\ \text{m/s}$
Substitute and take the root.
Check with work and energy.
$W = \displaystyle\int_0^3 8x\,dx = 36\ \text{J} = \tfrac{1}{2} \times 2.0 \times (40 - 4.0)$
The work equals the kinetic energy gained.
Integrate the acceleration.
$v = \displaystyle\int_0^t6t'^2\,dt' = 2t^3$
The force depends on time.
Substitute the time.
$v = 2 \times 2^3$
At $t = 2$ s.
Evaluate the velocity.
A bead on a wire feels a net force $F = -4x^3$ that depends only on its position. Which form of Newton's second law lets you find its speed at each position?
Complete the worked solution: a $3$ kg block starts from rest on a frictionless floor and is pushed with $F = 36t$ (newtons, seconds). Find its acceleration in m/s², velocity in m/s and position in m at $t = 4$ s.
Divide the force at that time by the mass.
$a = \dfrac{F}{m} =$ a
Newton's second law at the instant.
Integrate the acceleration from zero to the time.
$v = \displaystyle\int_0^ta\,dt' =$ v
From rest.
Integrate the velocity from zero to the time.
$x = \displaystyle\int_0^tv\,dt' =$ x
From the origin.
Check the powers of time.
$a \propto t, \quad v \propto t^2, \quad x \propto t^3$
Each integration raises the power by one.
Match each kind of net force to the way the second law is solved.
| integrate $F(t)/m$ over time | separate: $m\,dv/F(v) = dt$ | $mv\,dv = F(x)\,dx$ | constant-acceleration equations | |
|---|---|---|---|---|
| force of time | ||||
| force of velocity | ||||
| force of position | ||||
| constant force |
A $2$ kg cart starts from rest on a frictionless track and is pushed with a force that grows steadily, $F = 24t$ (newtons, seconds). Fill in its acceleration, velocity and position at $t = 1$ s.
| value | |
|---|---|
| acceleration (m/s²) | |
| velocity (m/s) | |
| position (m) |
A $1$ kg puck starts from rest and feels a net force $F = 7 - 4t$ (newtons, seconds). Write its velocity $v$ as a formula in $t$, in m/s.
Answer:
A particle moving along the $x$ axis has acceleration $a = 6x^2$ (m/s², with $x$ in meters). It passes $x = 0$ at $6$ m/s. What is the square of its speed at $x = 1$ m, in m²/s²?
Answer: m²/s² of speed squared
A $1000$ kg rocket sled on the test track at Holloman Air Force Base in New Mexico starts from rest. Its thrust starts at $40$ kN and falls steadily to zero over $6$ s. Ignoring friction and drag, how fast is it moving when the thrust ends, in m/s?
Answer: m/s at burnout
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $1$ kg puck starts from rest and feels a net force $F = 4 - 4t$ (newtons, seconds). Write its velocity $v$ as a formula in $t$, in m/s.
Answer:
You can solve Newton's second law when the force changes. Explain to someone why $x = \tfrac{1}{2}at^2$ gives the wrong answer for a spring.
19. Your turn: a $1.0$ kg particle starts from rest with $F = 6t^2$ (N, s). Find its velocity at $t = 2$ s., step 3
$v = 16\ \text{m/s}$
Starting from rest.