Analysis of uniform rectilinear motion graphs

Data

• The data of the problem are implicit in the graph

Preliminary considerations

• The general expression of uniform linear motion is given by $x=x_0+v·t$ . We will look for x₀ and v in each segment

• The term accompanying t corresponds to the velocity of the body according to the general expression. Don't forget that the instantaneous velocity of a body is defined as the derivative of the position with respect to time, so:

  $$v=\frac{dx}{dt}$$

  therefore, the velocity is the slope of each straight line in the graph position-time ( x-t )

• Remember that the position, velocity and acceleration are vector quantities. In the event that the trajectory is a straight line, we can use the sign convention in rectilinear motion and use scalars (numbers) instead of vectors

Resolution

There are 5 segments, each one with its own expression for its u.r.m.

Segment I ( 0 < t < 25 )

$$\begin{gathered} v_1=\frac{∆x}{∆t}=\frac{25-0}{25-0}=1 m/s \\ {x}_{01}=0 m \end{gathered}$$

The equation of motion in this segment is:

$$x_1=t$$

Segment II ( 25 < t < 37.5   )

$$v_2=\frac{∆x}{∆t}=\frac{75-25}{37.5-25}=4m/s$$

Now, to determine x₀₂ we can use the value of the straight line at any instant of time, e.g. t = 37.5 s, obtaining:

$$75=x_{02}+4·37.5\Rightarrow x_{02}=75-4·37.5=-75 m$$

The equation of motion in this second segment:

$$x_2=-75+4·t$$

Segment III ( 37.5 < t < 50 )

The body is stationary in this section, since there is no variation of x(t) over time. This means that:

$$\begin{gathered} v_3=0 m/s \\ {x}_{03}=75 m \end{gathered}$$ 

The equation of motion for the third segment:

$$x_3=75$$

Segment IV ( 50 < t < 75 )

The slope (velocity) on this segment is equal to the slope of the first segment but in the opposite direction. Following the sign convention for rectilinear motion from above, this means that the motion has changed its direction and is directed to the starting point. Let's see:

$$v_4=\frac{∆x}{∆t}=\frac{50-75}{75-50}=-1 m/s$$

Now, to determine x₀₄ we can use the value of the straight line at any time, e.g. t = 50 s, resulting in:

$$75=x_{04}-1·50\Rightarrow x_{04}=75+50=125 m$$

For the fourth segment, we have the equation of motion:

$$x_4=125-t$$

Segment V ( 75 < t < 87.5 ) 

Again, the slope in this segment (velocity) is like in the second segment but in opposite direction. Following the sign convention for rectilinear motion from above, this means that the motion has changed its direction and is directed to the starting point. Let's see:

$$v_5=\frac{∆x}{∆t}=\frac{0-50}{87.5-75}=-4m/s$$

We calculate the initial position using as reference the final point on the graph:

$$0=x_{05}-4·87.5\Rightarrow x_{05}=350 m$$

Finally, for the 5th segment the equation of motion is:

$$x_5=350-4·t$$

Once obtained the expression for each section, we can write a single expression in the form of a piecewise-defined function:

$$x\left(t\right)=\left\{\begin{array}{c}\begin{array}{c}t 0<t<25\\ -75+4·t 25<t<37.5\\ 75 37.5<t<50\end{array}\\ 125-t 50<t<75\\ 350-4·t 75<t<87.5\end{array}\right.$$ 

And for the velocity...

$$v\left(t\right)=\left\{\begin{array}{c}\begin{array}{c}1 0<t<25\\ 4 25<t<37.5\\ 0 37.5<t<50\end{array}\\ -1 50<t<75\\ -4 75<t<87.5\end{array}\right.$$