Find the object's positions \(\texttip{x_{\rm 1}}{x_1}\), \(\texttip{x_{\rm 2}}{x_2}\),at times \(t_1=2.0\; \rm s\), \(t_2=4.0\; \rm s\), \(t_3=13\; \rm s\), and \(t_4=17\; \rm s\).

Question

A continuous curve that shows an object's position as a function of time is called a position-versus-time graph. You can gain relevant information on an object's motion if you interpret such graphs correctly. To do that, it is useful to review the definition of velocity. For motion along a line, the velocity \(\texttip{v_{\mit x}}{v_x}\) is the ratio of the displacement \(\texttip{\Delta x}{Deltax}\) of an object to the time interval \(\texttip{\Delta t}{Deltat}\) during which this displacement occurs, which can be written as

\(\large{v_x=\frac{\Delta x}{\Delta t}}\).

This equation has a graphical interpretation: It tells us that \(\texttip{v_{\mit x}}{v_x}\) is the slope of the position-versus-time graph representing the motion. This implies that you can associate the slope of the graph, a geometrical quantity, with the physical quantity velocity. This and other aspects of interpreting position-versus-time graphs are outlined in the following Tactics Box.

Answer 1

To work out velocity, position or time, the relationship V=dx/dt or V=∂x/∂t where ∂x is the difference in position (in a specific direction), and ∂t the corresponding difference in time.

t1=2.0, t2=4.0, t3=13 and t4=17 give 4 time values, but position (x) values are needed for the positions.

A function like x(t)=a0+a1t+a2t^2+a3t^3+... etc is needed to relate time and position.

Alternatively we need a velocity function v(t) from which position can be calculated using integration and initial conditions. x=∫vdt between specific limits for t.