Write the position of a point moving on a number line as a function of time, x(t), and the slope of the tangent to the position-time graph at any moment is the velocity at that moment.
Differentiate position once with respect to time and you get velocity; differentiate velocity once more and you get acceleration.
When the velocity is 0 the point stops for an instant, but it only turns around if the sign of the velocity changes there.
Move the time t to see the slope of the tangent together with the direction in which point P travels on the number line at the right.
Differentiate Position to Get Velocity
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Graph of the position x(t) = t³ − 6t² + 9t with a tangent; at the right, point P on the number line
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👀 What to read in the picture
①The blue curve is the position x(t) over time, and point P on the number line at the right is the point that actually moves
②The slope of the tangent is the velocity v at that moment. A green point means v > 0 and P moves up, in the positive direction; an orange point means v < 0 and P moves down
③At t = 1 and t = 3, v = 0 and the sign of v changes, so P turns around. Its position is 4 at t = 1 and 0 at t = 3
Position, Velocity and Acceleration
Velocity is the derivative of position
v(t) = dxdt = x′(t)
Differentiating the position x(t) with respect to time gives the velocity v(t)
Acceleration is the derivative of velocity
a(t) = dvdt = x′′(t)
Differentiating velocity once more gives acceleration, the same as differentiating position twice
A Change of Sign Means a Change of Direction
When the direction changes
v(t) = 0 and v changes sign there ⇒ the direction changes at that time
v = 0 alone is not enough: the sign has to actually change for the direction to change
🔄 The sign of v is the direction
①If v > 0 the point moves in the positive direction; if v < 0, in the negative direction
②If v = 0 at an instant and its sign changes there, the direction of motion changes at that instant
③Speed is the absolute value of velocity, |v|; at an instant where the acceleration a = 0, the velocity can have an extreme value
Worked Examples
Example 1
Point P has position x(t) = t³ − 6t² + 9t. Find its velocity and acceleration at t = 2.
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Differentiate the position once and twice to get velocity and acceleration.
v(t) = 3t² − 12t + 9, a(t) = 6t − 12
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Substitute t = 2.
v(2) = 12 − 24 + 9 = −3, a(2) = 12 − 12 = 0
▸ Velocity −3, acceleration 0
The velocity is negative, so P is moving in the negative direction. Set t = 2 in the picture to see the same values.
Example 2
Find every time at which the same point P changes direction (t > 0).
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Solve v(t) = 0.
v(t) = 3(t − 1)(t − 3) = 0 ⇒ t = 1, t = 3
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v changes sign around both t = 1 and t = 3.
▸ t = 1, t = 3
Only after checking for a sign change at each time where v = 0 can you call it a change of direction.