1.16 Rectilinear motion (application)

Problem : Two boys (P and Q) walk to their respective school from their homes in the morning on a particular day. Their motions are plotted on a position – time graph for the day as shown. If all schools and homes are situated by the side of a straight road, then answer the followings :

• Which of the two resides closer to the school?
• Which of the two starts earlier for the school?
• Which of the two walks faster?
• Which of the two reaches the school earlier?
• Which of the two overtakes other during walk?

Solution : In order to answer questions, we complete the drawing with vertical and horizontal lines as shown here. Now,

1: Which of the two resides closer to the school?

We can answer this question by knowing the displacements. The total displacements here are OC by P and OD by Q. From figure, OC<OD. Hence, P resides closer to the school.

2: Which of the two starts earlier for the school?

The start times are point O for P and A for Q on the time axis. Hence, P starts earlier for school.

3: Which of the two walks faster?

The speed is given by the slope of the plot. Slope of the motion of P is smaller than that of Q. Hence, Q walks faster.

4: Which of the two reaches the school earlier?

Both boys reach school at the time given by point B on time axis. Hence, they reach school at the same time.

5: Which of the two overtakes other during walk?

The plots intersect at a point E. They are at the same location at this time instant. Since, speed of B is greater, he overtakes A.

Problem : A displacement – time plot in one dimension is as shown. Find the ratio of velocities represented by two straight lines.

Solution : The slope of displacement – time plot is equal to velocity. Let v1 and v2 be the velocities in two segments, then magnitudes of velocities in two segments are :

$$\begin{array}{l}\left|v_1\right|=\tan 60°=\surd 3\\ \left|v_2\right|=\tan 30°=\frac{1}{\surd 3}\end{array}$$

We note that velocity in the first segment is positive, whereas velocity in the second segment is negative. Hence, the required ratio of two velocities is :

$$\begin{array}{l}\frac{v_1}{v_2}=-\frac{\surd 3}{\frac{1}{\surd 3}}=-3\end{array}$$

Problem : The displacement “x” of a particle moving in one dimension is related to time “t” as :

$$\begin{array}{l}t=\surd x+3\end{array}$$

where “t” is in seconds and “x” is in meters. Find the displacement of the particle when its velocity is zero.

Solution : We need to find “x” when velocity is zero. In order to find this, we require to have an expression for velocity. This, in turn, requires an expression of displacement in terms of time. The given expression of time, therefore, is required to be re-arranged :

$$\begin{array}{l}\surd x=t-3\end{array}$$

Squaring both sides, we have :

$$\begin{array}{l}\Rightarrow x={(t-3)}^2=t^2-6t+9\end{array}$$

Now, we obtain the required expression of velocity in one dimension by differentiating the above relation with respect to time,

$$\begin{array}{l}v=\frac{dx}{dt}=2t-6\end{array}$$

According to question,

$$\begin{array}{l}v=2t-6=0\\ t=3s\end{array}$$

Putting this value of time in the expression of displacement, we have :

$$\begin{array}{l}x=3^2-6x3+9=0\end{array}$$

Problem : A particle moving in a straight line covers 1/3rd of the distance with a velocity 4 m/s. The remaining part of the linear distance is covered with velocities 2 m/s and 6 m/s for equal times. What is the average velocity during the total motion?

Solution : The average velocity is ratio of displacement and time. In one dimensional unidirectional motion, distance and displacement are same. As such, average velocity is ratio of total distance and time.

Let the total distance be “x”. Now, we need to find total time in order to find the average velocity. Let " $t_1$ ", " $t_2$ " and " $t_2$ " be the time periods of the motion in three parts of the motion as given in the question. Considering the first part of the motion,

$$\begin{array}{l}{t}_1=\frac{x_1}{v_1}=\frac{\frac{x}{3}}{4}=\frac{x}{12}\end{array}$$

For the second part of the motion, the particle covers the remaining distance at two different velocities for equal time, “ $t_2$ ”,

$$\begin{array}{l}x-\frac{x}{3}=v_{2}x{t}_2+v_{3}x{t}_2\\ \Rightarrow \frac{2x}{3}=2t_2+6t_2=8t_2\\ \Rightarrow t_2=\frac{2x}{24}=\frac{x}{12}\end{array}$$

Thus, average velocity is :

$$\begin{array}{l}{v}_a=\frac{x}{t}=\frac{x}{t_1+2t_2}=\frac{x}{\frac{x}{12}+\frac{x}{6}}=4m/s\end{array}$$