Velocity and Acceleration Calculator

This velocity and acceleration calculator takes a position function \( s(t) \) and a moment in time, and returns the velocity \( v(t) = s'(t) \), the acceleration \( a(t) = s''(t) \), their values at that moment, and whether the object is speeding up or slowing down.

It’s aimed at AP Calculus and physics students working through straight-line motion (“particle moving along a line”) problems. It’s free, runs in your browser and needs no sign-up.

How to use the velocity and acceleration calculator

1. Type the position function into s(t) using the variable t. Use ^ for powers (t^3), and sin(), cos(), sqrt(), ln() and e^(t) as needed. Implicit multiplication works: 9t^2 and t sin(t) are fine.
2. Enter the time you care about in At time t. Values like pi/2 are accepted.
3. Press Analyze motion. You’ll see the formulas for \( v(t) \) and \( a(t) \), then \( s \), \( v \) and \( a \) at your time, plus a one-line verdict: speeding up, slowing down, momentarily at rest, or constant velocity at that instant.

How it works

Velocity is the rate of change of position and acceleration is the rate of change of velocity:

$$v(t) = s'(t), \qquad a(t) = v'(t) = s''(t)$$

The calculator differentiates \( s(t) \) symbolically, twice, and evaluates all three functions at your \( t \). For polynomials, that’s just the power rule applied term by term.

The speeding-up test compares signs. Speed is \( |v| \), and it grows when velocity and acceleration point the same way:

• \( v \) and \( a \) have the same sign: speeding up.
• \( v \) and \( a \) have opposite signs: slowing down.
• \( v = 0 \): the object is momentarily at rest (a possible turnaround).

Worked example

A particle moves along a line with \( s(t) = t^3 - 9t^2 + 24t \) meters, for \( t \) in seconds. Describe its motion at \( t = 5 \) and at \( t = 1 \). You can enter the same function in the calculator above.

Derivatives:

$$v(t) = 3t^2 - 18t + 24, \qquad a(t) = 6t - 18$$

At \( t = 5 \): \( s(5) = 125 - 225 + 120 = 20 \) m, \( v(5) = 75 - 90 + 24 = 9 \) m/s, \( a(5) = 12 \) m/s². Both are positive, so the particle is moving right and speeding up.

At \( t = 1 \): \( s(1) = 16 \) m, \( v(1) = 9 \) m/s, \( a(1) = -12 \) m/s². Velocity is positive but acceleration is negative, so the particle is moving right while slowing down.

When is it at rest? Factor the velocity: \( v(t) = 3(t - 2)(t - 4) \), so \( v = 0 \) at \( t = 2 \) and \( t = 4 \). These are the critical points of \( s \), where the particle can change direction. It moves right until \( t = 2 \) (reaching \( s = 20 \)), left until \( t = 4 \) (back to \( s = 16 \)), then right again.

Displacement vs total distance

Exams love this follow-up, and the calculator won’t do it for you, so here is how. Over \( 0 \le t \le 5 \):

• Displacement is \( s(5) - s(0) = 20 - 0 = 20 \) m. Equivalently, \( \int_0^5 v(t)\,dt = 20 \) by the Fundamental Theorem of Calculus.
• Total distance adds up every leg without cancellation: \( 20 + 4 + 4 = 28 \) m, which equals \( \int_0^5 |v(t)|\,dt \). The definite integral calculator can evaluate that integral if you type the velocity with x in place of t inside abs().

Use the turnaround times from \( v(t) = 0 \) to split the interval. Use the calculator at each endpoint and turnaround time to read off the positions.

Common mistakes

• Reading “negative acceleration” as “slowing down.” At \( t = 3.5 \), \( a = 3 > 0 \) but \( v = -2.25 < 0 \), so the particle slows down even though \( a \) is positive.
• Forgetting the units. If \( s \) is in meters and \( t \) in seconds, \( v \) is in m/s and \( a \) in m/s².
• Using \( x \) instead of \( t \). The input expects the variable t.

Related guides

• Power Rule for Derivatives
• How to Find Critical Points
• Fundamental Theorem of Calculus
• Related Rates

Further reading

• Tangent Lines and Rates of Change (Paul’s Online Math Notes) — velocity as the instantaneous rate of change of position.
• Higher Order Derivatives (Paul’s Online Math Notes) — second derivatives, including acceleration as \( s''(t) \).

FAQ

How do you find velocity from a position function?

Differentiate it: \( v(t) = s'(t) \). Enter \( s(t) \) in the calculator and it shows \( v(t) \) as a formula and its value at your chosen time.

How do you find acceleration from position?

Differentiate twice: \( a(t) = s''(t) \). The calculator displays \( a(t) \) and evaluates it at your time.

How can I tell if an object is speeding up?

Check the signs of \( v \) and \( a \) at that instant. Same sign means speeding up; opposite signs mean slowing down. The calculator prints this verdict for you.

Can it go from acceleration back to position?

No. It only works forward, from \( s(t) \) to \( v(t) \) and \( a(t) \). Going backward requires integration plus initial conditions; the integral calculator handles the antiderivative step.