Velocity & Acceleration

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 means zero velocity — it pauses and reverses direction

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

2020 CSAT Math (Calculus) type, adapted

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

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