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Differentiate for Velocity and Acceleration, Integrate for Distance

When a question gives velocity or position as a formula instead of a constant acceleration, the equations of motion no longer apply. You use calculus: differentiate for the next quantity, integrate for the previous one.

NSEJS · Kinematics · 6 linked questions · 100% coverage

Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction

Study deeper: NCERT Science · Kinematics

The relations

v = dx/dt, a = dv/dt, and by the chain rule a = v dv/dx whenever velocity is given as a function of position

Distance between two instants is the integral of speed over that interval, and displacement is the integral of velocity.

For v = A t + B t², the distance travelled between t = 2 s and t = 3 s is the integral from 2 to 3 of (A t + B t²) dt = (5/2) A + (19/3) B

Velocity as a function of position

If v = β x^(-4n), then a = v dv/dx = β x^(-4n) x (-4n β x^(-4n - 1)) = -4n β² x^(-8n - 1)

If v = α sqrt(x), then dx/dt = α sqrt(x) gives sqrt(x) = α t / 2, so x = α² t² / 4, and v = α² t / 2 grows in proportion to t: the acceleration is constant, even though the velocity was given in terms of x.

Particle that comes to rest and returns

For x = 50 + 12 t - t³, the velocity is v = 12 - 3 t², which is zero at t = 2 s

The position then is 50 + 24 - 8 = 66 m, so the particle moves from 50 m to 66 m, a distance of 16 m, before coming to rest. When a particle reverses direction inside an interval, split the integral at the turning point and add the magnitudes to get the distance.

Powers of t and the relation to v

If y = (t + 10)^(-1), then the velocity is v = -(t + 10)^(-2) and the acceleration is a = 2 (t + 10)^(-3)

Since |v|^(3/2) = (t + 10)^(-3), the acceleration is proportional to (velocity)^(3/2). To express one quantity in terms of another, eliminate t by writing both as powers of the same expression.

OlympIQ exam lens

✓v = dx/dt, a = dv/dt = v dv/dx
✓Distance between two times = integral of speed; displacement = integral of velocity
✓If v is given as a function of x, use a = v dv/dx
✓Find the turning point (v = 0) before computing distance if the direction reverses
✓To relate two quantities, write both as powers of one expression and compare exponents

OlympIQ trap

Do not use v = u + at or s = ut + (1/2)at² when the acceleration varies with time or position. Those formulas need constant acceleration; here you must integrate.

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