2.6 One dimensional motion with constant acceleration  (Page 6/6)

This is a case of one dimensional motion with constant acceleration. Since both cyclists cross the finish line simultaneously, they cover same displacement in equal times. Hence,

$$\begin{array}{l}{x}_1=x_2\end{array}$$

$$\begin{array}{l}{u}_{1}t+\frac{1}{2}{a}_1{t}^2=u_{2}t+\frac{1}{2}{a}_2{t}^2\end{array}$$

Putting values as given in the question, we have :

$$\begin{array}{l}2t+\frac{1}{2}x2x{t}^2=4t+\frac{1}{2}x1x{t}^2\\ \Rightarrow t^2-4t=0\\ \Rightarrow t=0s,4s\end{array}$$

Neglecting zero value,

$$\begin{array}{l}\Rightarrow t=4s\end{array}$$

The linear distance covered by the cyclist is obtained by evaluating the equation of displacement of any of the cyclists as :

$$\begin{array}{l}x=2t+\frac{1}{2}x2x{t}^2=2x4+\frac{1}{2}x2x{4}^2=24t\end{array}$$

Both cars start from rest. They move with different accelerations and hence take different times to reach equal distance, say $t_1$ and $t_2$ for cars A and B respectively. Their final speeds at the end points are also different, say $v_1$ and $v_2$ for cars A and B respectively. According to the question, the difference of time is “t”, whereas difference of speeds is “v”.

As car A is faster, it takes lesser time. Here, $t_2>t_1$ . The difference of time, “t”, is :

$$\begin{array}{l}t=t_2-t_1\end{array}$$

From equation of motion,

$$\begin{array}{l}x=\frac{1}{2}{a}_1{t_1}^2\\ \Rightarrow t_1=\surd \left(\frac{2x}{a_1}\right)\end{array}$$

Similarly,

$$\begin{array}{l}\Rightarrow t_2=\surd \left(\frac{2x}{a_2}\right)\end{array}$$

Hence,

$$\begin{array}{l}\Rightarrow t=t_2-t_1=\surd \left(\frac{2x}{a_2}\right)-\surd \left(\frac{2x}{a_1}\right)\end{array}$$

Car A is faster. Hence, $v_1>v_2$ . The difference of time, “v”, is :

$$\begin{array}{l}\Rightarrow v=v_1-v_2\end{array}$$

From equation of motion,

$$\begin{array}{l}\Rightarrow {v_1}^2=2a_{1}x\\ \Rightarrow v_1=\surd \left(2a_{1}x\right)\end{array}$$

Similarly,

$$\begin{array}{l}\Rightarrow v_2=\surd \left(2a_{2}x\right)\end{array}$$

Hence,

$$\begin{array}{l}v=v_1-v_2=\surd \left(2a_{1}x\right)-\surd \left(2a_{2}x\right)\end{array}$$

The required ratio, therefore, is :

$$\begin{array}{l}\frac{v}{t}=\frac{\surd \left(2a_{1}x\right)-\surd \left(2a_{2}x\right)}{\surd \left(\frac{2x}{a_2}\right)-\surd \left(\frac{2x}{a_1}\right)}=\frac{\{\surd (2a_1)-\surd (2a_2\left)\right\}\surd a_1\surd a_2}{\{\surd (2a_1)-\surd (2a_2\left)\right\}}\end{array}$$

$$\begin{array}{l}\frac{v}{t}=\surd \left(a_1{a}_2\right)\end{array}$$

One of the particles begins with a constant velocity, “v” and continues to move with that velocity. The second particle starts with zero velocity and continues to move with a constant acceleration, “a”. At a given instant, “t”, the first covers a linear distance,

$$\begin{array}{l}{x}_1=vt\end{array}$$

In this period, the second particle travels a linear distance given by :

$$\begin{array}{l}{x}_2=\frac{1}{2}a{t}^2\end{array}$$

First particle starts with certain velocity as against second one, which is at rest. It means that the first particle will be ahead of second particle in the beginning. The separation between two particles is :

$$\begin{array}{l}\Delta x=x_1-x_2\\ \Delta x=vt-\frac{1}{2}a{t}^2\end{array}$$

For the separation to be maximum, its first time derivative should be equal to zero and second time derivative should be negative. Now, first and second time derivatives are :

$$\begin{array}{l}\frac{đ(\Delta x)}{đt}=v-at\\ \frac{{đ}^2(\Delta x)}{đt^2}=-a<0\end{array}$$

For maximum separation,

$$\begin{array}{l}\Rightarrow \frac{đ(\Delta x)}{đt}=v-at=0\end{array}$$

$$\begin{array}{l}\Rightarrow t=\frac{v}{a}\end{array}$$

The separation at this time instant,

$$\begin{array}{l}\Delta x=vt-\frac{1}{2}a{t}^2=vx\frac{v}{a}-\frac{1}{2}a{\left(\frac{v}{a}\right)}^2\\ \Delta x=\frac{v^2}{2a}\end{array}$$