Velocity and Displacement by Integration

The chain run backwards

Differentiating position s with respect to time gives velocity v, and differentiating v gives acceleration a. For $s=t^2$s = t², v = 2t and a = 2.

Integrating undoes each step: integrate a to get v, then integrate v to get s. Each integration brings a constant, and since a constant differentiates to 0, the rate alone cannot fix it. It is fixed by one known value, usually the value at the start.

A ball thrown upward at 20 meters per second, with a = −10 throughout. Integrating a = −10, with v = 20 at t = 0, gives v = 20 − 10t; integrating again, with s = 0 at t = 0, gives $s=20t-5t^2$s = 20t − 5t². At t = 1 the three graphs read s = 15, v = 10 and a = −10. Drag t to 2, where v = 0 and the ball is at its highest, s = 20.

The start pins the constant

v = 2t integrates to $s=t^2+C$s = t² + C. If the particle is at s = 0 when t = 0, then 0 = 0 + C, so C = 0 and $s=t^2$s = t². If it is at s = 5 when t = 0, then C = 5 and $s=t^2+5$s = t² + 5: the same motion, begun 5 meters further along.

Starting from an acceleration takes two constants. Take a = 6t − 4, with v = 3 and s = 1 when t = 0. Integrating a gives $v=3t^2-4t+C$v = 3t² − 4t + C, and v = 3 at t = 0 makes C = 3, so $v=3t^2-4t+3$v = 3t² − 4t + 3.

Integrating v gives $s=t^3-2t^2+3t+D$s = t³ − 2t² + 3t + D, and s = 1 at t = 0 makes D = 1, so $s=t^3-2t^2+3t+1$s = t³ − 2t² + 3t + 1. At t = 2 the position is 8 − 8 + 6 + 1 = 7.

Displacement is a definite integral

The integral of v from $t=t_1$t = t₁ to $t=t_2$t = t₂ is $s\left(t_2\right)-s\left(t_1\right)$s(t₂) − s(t₁), the change in position, called the displacement. The constant appears in both terms and cancels, so it is not needed.

For $v=3t^2$v = 3t² from t = 0 to 2, $s=t^3$s = t³, so the displacement is $2^3-0^3=8$2³ − 0³ = 8. From t = 1 to 2 it is $2^3-1^3=7$2³ − 1³ = 7.

The final position is the starting position plus the displacement. For $v=3t^2-4t+3$v = 3t² − 4t + 3 from 0 to 2, the displacement is (8 − 8 + 6) − 0 = 6, and the particle started at s = 1, so it ends at 1 + 6 = 7, as found above.

Under $v=3t^2$v = 3t² from t = 0 to 2 the shaded area is 8, the displacement. The velocity reaches 12 at t = 2, but the particle started from rest, so it moved only 8.

A car speeds up for 2 seconds, holds 6 meters per second, then brakes to rest at t = 6. The area under its velocity graph is the distance it has moved, filled into the tank. At t = 3 the area is 6 + 6 = 12, so s = 12 meters. Drag t on to 6, and s stops at 24 meters.

When v is negative

Where the velocity graph is below the axis the particle moves backward, and that stretch of the integral counts negative. For v = 2 − t, the particle moves forward until t = 2 and backward after it.

Integrating, $s=2t-\frac{t^2}{2}$s = 2t − t²/2, and from t = 0 to 4 the displacement is (8 − 8) − 0 = 0: the particle ends where it began. Yet each triangle between the graph and the axis has area ½ × 2 × 2 = 2, so it covered 2 + 2 = 4 meters of ground. The signed integral gives the displacement, 0; the total distance adds the sizes of the pieces, 4.

v = 2 − t, forward from t = 0 to 2 and backward from 2 to 4, turning at the dot. The triangle above the axis counts +2 and the one below counts −2, so the displacement is 0 and the distance is 4.

The usual mistakes

Giving the velocity at the end. For v = 2t, the velocity at t = 3 is 6, but the displacement from 0 to 3 is $3^2=9$3² = 9.

Multiplying the final velocity by the time. 6 × 3 = 18 treats the particle as moving at 6 the whole time; it started from rest, and the area under v = 2t is the triangle ½ × 3 × 6 = 9.

Integrating the power wrongly. $3t^2$3t² integrates to $t^3$t³, so from 0 to 2 the displacement is 8, not $2^2=4$2² = 4.

Leaving out the constant. $s=t^2+C$s = t² + C gives the position only once C is fixed by the start.

A survey vehicle on a tether

In the application below, a vehicle hanging from a ship rises, sinks and rises again. Its final depth is its starting depth less its displacement upward, from one integral of v. The tether the winch moves is a total distance, which needs the integral split at the times where v = 0.