high school Velocity & Acceleration

Velocity & Acceleration

On a position-versus-time graph, the slope of the tangent at any instant is exactly the velocity at that moment. So differentiating position once gives velocity, and differentiating velocity once more gives acceleration. When the slope reaches zero the object pauses and may reverse, but a real turnaround happens only when the velocity actually changes sign. Drag the time t slider and the tangent slope and the dot's color change, and the velocity becomes zero at the turning point.

Differentiate Position to Get Velocity

Time t2

👀 See It

①The curve is position x(t) over time

②The tangent slope at an instant is the velocity v then

③Zero slope is v = 0. Reverse only if the sign changes

④Accel is the box below. a = 0 is a velocity extremum

Position, Velocity, Acceleration

Velocity = derivative of position

v(t) = dxdt = x′(t)

Differentiating position x(t) in time gives velocity v(t)

Acceleration = derivative of velocity

a(t) = dvdt = x′′(t)

Differentiating velocity once more gives acceleration — the second derivative of position

Zero Velocity Means a Turn

Change of direction

v(t)=0 and v changes sign across t ⇒ direction reverses at that time

v=0 alone is not enough — the sign must actually flip

🔄 The Sign of v Is the Direction

①v>0 moves in the positive direction, v<0 in the negative

②At the instant v=0 the direction reverses (when the sign flips)

③Speed is the absolute value |v|

Compute It Directly

Example 1

A point P has position x(t)=t³−6t²+9t. Find its velocity and acceleration at t=2.

Take the first and second derivatives for velocity and acceleration.

v(t)=3t²−12t+9, a(t)=6t−12

Substitute t=2.

v(2)=12−24+9=−3, a(2)=12−12=0

▸ velocity −3, acceleration 0

A negative velocity means it is currently moving in the negative direction.

Example 2

Find every time the same point P changes direction. (t>0)

Solve v(t)=0.

v(t)=3(t−1)(t−3)=0 ⇒ t=1, t=3

Across both t=1 and t=3 the sign of v flips.

▸ t = 1, t = 3

At each time v=0, you must confirm a sign change for a genuine turn.

Wrap-up

Key relationship

position x(t) → differentiate → velocity v(t) → differentiate → acceleration a(t)

One derivative gives velocity, two give acceleration

exam-style

A point P on a number line has position x(t)=t³−3t². At what time does P change direction? (t>0)

①t = 0

②t = 1

③t = 2

④t = 3

⑤It never changes direction

▸ ③ t = 2

Find the velocity.

v(t)=3t²−6t=3t(t−2)

For t>0, the sign of v flips from negative to positive at t=2.

direction reverses at t = 2

🎯 Exam Points

①v=x′, a=v′=x″ (one and two derivatives)

②v>0 positive direction, v<0 negative

③A turn needs v=0 AND a sign change

④Speed is |v|

⑤t=0 is the start, usually excluded from direction changes

← Previous

Equations & Inequalities

Integration Methods

Was this helpful? Support seegongsik