# Kinematics Graphs

The a-t, v-t and s-t graphs are inter-related:

• The gradient of the st graph is v.
• $\frac{ds}{dt}=v \quad$
• The gradient of the vt graph is a.
• $\frac{dv}{dt}=a\quad$
• The area under the at graph is Δv.
• $\int_{{{t}_{1}}}^{{{t}_{2}}}{a}\,dt={{v}_{2}}-{{v}_{1}}$
• The area under the vt graph is Δs.
• $\int_{{{t}_{1}}}^{{{t}_{2}}}{v}\,dt={{s}_{2}}-{{s}_{1}}$

Example

Given the v-t graph for the motion of a ball over 5 seconds, derive the corresponding s-t and a-t graphs. (Assume “rightward is positive” sign convention.) Solution • The areas of +1.0, +0.5, -0.5 and -1.0 in the v-t graph correspond to the s-t graph increasing by 1.0 m and 0.5 m, before decreasing by 0.5 m and 1.0 m.
• As the ball continues to move rightward until t = 2 s, the maximum displacement occurs at t = 2 s.
• The ball returns to the starting point after 5 s because the positive area happens to match the negative area exactly. • As the velocity changes, so does the gradient of the v-t
• Constant speed segments in v-t graphs translate into straight lines in s-t graphs.
• Increasing speed segments translate into steepening curves, whereas decreasing speed segments translate into flattening curves. • The a-t graph is readily obtained from the gradient of the v-t • The areas of 0, -2.0 and +1.0 in the a-t graph correspond to velocity remaining constant, before decreasing by 2.0 m s-1 and then increasing by 1.0 m s-1.