seegongsik
Saved words
Grade 12 / High 3 (age 17-18)

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 to watch how the tangent slope and the dot's color change, and notice exactly when the velocity becomes zero.

Differentiate Position to Get Velocity
2
👀 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.
1
Take the first and second derivatives for velocity and acceleration.
v(t)=3t²−12t+9, a(t)=6t−12
2
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)
1
Solve v(t)=0.
v(t)=3(t−1)(t−3)=0 ⇒ t=1, t=3
2
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)
t = 0
t = 1
t = 2
t = 3
It never changes direction
③ t = 2
1
Find the velocity.
v(t)=3t²−6t=3t(t−2)
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
Next →
Integration Methods
Was this helpful? Support seegongsik