Gradient of s–t=v of v–t=a s(t_2)-s(t_1)= _t_1t_2v, dt Example Displacement-time graph: gradient = velocity. Velocity-time graph: gradient = acceleration; signed area under graph = change in displacement. Displacement–time graph For 0<t<2, v= 6-02-0=3. For 2<t<5, the graph is horizontal, so v=0. For 5<t<8, v= -6-68-5=-4. Velocity–time graph For 0<t<2, a= 4-02-0=2. For 2<t<5, the graph is horizontal, so a=0. For 5<t<8, a= -2-48-5=-2. From t=0 to t=8, change in displacement = 12(3+7)(4)- 12(1)(2) =20-1 =19 total distance=20+1=21 Below the time axis, area contributes negative displacement but positive distance. Position relative to the origin also requires the starting position. A two-sided derivative does not exist at a corner; an endpoint may have a one-sided rate.

\(\text{Gradient of }s\text{–}t=v\qquad\text{Gradient of }v\text{–}t=a\)

\(s(t_2)-s(t_1)=\int_{t_1}^{t_2}v\,\mathrm dt\)

For \(2<t<5\), the graph is horizontal, so \(v=0\).

For \(5<t<8\), \(v=\dfrac{-6-6}{8-5}=-4\).

For \(2<t<5\), the graph is horizontal, so \(a=0\).

For \(5<t<8\), \(a=\dfrac{-2-4}{8-5}=-2\).

\(\text{change in displacement}\)\({}=\dfrac12(3+7)(4)-\dfrac12(1)(2)\)\({}=20-1\)\({}=19\)

Below the time axis, area contributes negative displacement but positive distance. Position relative to the origin also requires the starting position.

A two-sided derivative does not exist at a corner; an endpoint may have a one-sided rate.

Displacement-Time and Velocity-Time Graphs

Displacement-Time and Velocity-Time Graphs