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During upward motion, velocity and acceleration due to gravity are in opposite direction. As a result, velocity decreases till it achieves the terminal velocity of zero at the end of 3rd second. Note that displacement during the motion is increasing till the ball reaches the maximum height.
At the maximum height, the velocity of the ball is zero and is under the action of force due to gravity as always during the motion. As such, the ball begins moving in downward direction with the acceleration due to gravity. The directions of velocity and acceleration, in this part of motion, are same. Note that displacement with respect to point of projection is decreasing.
In the overall analysis of motion when initial velocity is against acceleration, parameters defining motion i.e initial velocity, final velocity and acceleration act along a straight line, but in different directions. As a consequence, displacement may either be increasing or decreasing during the motion. This means that magnitude of displacement may not be equal to distance. For example, consider the motion of ball from the point of projection, A, to maximum height, B, to point, C, at the end of 4 seconds. The displacement is 40 m, while distance is 45 + 5 = 50 m as shown in the figure below.
For this reason, average speed is not always equal to the magnitude of average velocity.
$$\begin{array}{l}s\ne \left|x\right|\end{array}$$
and
$$\begin{array}{l}\frac{\Delta s}{\Delta t}\ne \left|\frac{\Delta x}{\Delta t}\right|\end{array}$$
Two cyclists start off a race with initial velocities 2 m/s and 4 m/s respectively. Their linear accelerations are 2 and 1 $m/{s}^{2}$ respectively. If they reach the finish line simultaneously, then what is the length of the track?
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 :
$$\begin{array}{l}2t+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{t}^{2}=4t+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{t}^{2}\\ \Rightarrow {t}^{2}-4t=0\\ \Rightarrow t=0\phantom{\rule{2pt}{0ex}}s,4\phantom{\rule{2pt}{0ex}}s\end{array}$$
Neglecting zero value,
$$\begin{array}{l}\Rightarrow t=4\phantom{\rule{2pt}{0ex}}s\end{array}$$
The linear distance covered by the cyclist is obtained by evaluating the equation of displacement of any of the cyclists as :
$$\begin{array}{l}x=2t+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{t}^{2}=2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}4+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{4}^{2}=24\phantom{\rule{2pt}{0ex}}t\end{array}$$
Two cars are flagged off from the starting point. They move with accelerations ${a}_{1}$ and ${a}_{2}$ respectively. The car “A” takes time “t” less than car “B” to reach the end point and passes the end point with a difference of speed, “v”, with respect to car “B”. Find the ratio v/t.
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,
$$\begin{array}{l}\Rightarrow {{v}_{1}}^{2}=2{a}_{1}x\\ \Rightarrow {v}_{1}=\surd \left(2{a}_{1}x\right)\end{array}$$
Similarly,
$$\begin{array}{l}\Rightarrow {v}_{2}=\surd \left(2{a}_{2}x\right)\end{array}$$
Hence,
$$\begin{array}{l}v={v}_{1}-{v}_{2}=\surd \left(2{a}_{1}x\right)-\surd \left(2{a}_{2}x\right)\end{array}$$
The required ratio, therefore, is :
$$\begin{array}{l}\frac{v}{t}=\frac{\surd \left(2{a}_{1}x\right)-\surd \left(2{a}_{2}x\right)}{\surd \left(\frac{2x}{{a}_{2}}\right)-\surd \left(\frac{2x}{{a}_{1}}\right)}=\frac{\{\surd (2{a}_{1})-\surd (2{a}_{2}\left)\right\}\surd {a}_{1}\surd {a}_{2}}{\{\surd (2{a}_{1})-\surd (2{a}_{2}\left)\right\}}\end{array}$$
$$\begin{array}{l}\frac{v}{t}=\surd \left({a}_{1}{a}_{2}\right)\end{array}$$
Two particles start to move from same position. One moves with constant linear velocity, “v”; whereas the other, starting from rest, moves with constant acceleration, “a”. Before the second catches up with the first, what is maximum separation between two?.
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 :
$$\begin{array}{l}\frac{\u0111(\Delta x)}{\u0111t}=v-at\\ \frac{{\u0111}^{2}(\Delta x)}{\u0111{t}^{2}}=-a<0\end{array}$$
For maximum separation,
$$\begin{array}{l}\Rightarrow \frac{\u0111(\Delta x)}{\u0111t}=v-at=0\end{array}$$
$$\begin{array}{l}\Rightarrow t=\frac{v}{a}\end{array}$$
The separation at this time instant,
$$\begin{array}{l}\Delta x=vt-\frac{1}{2}a{t}^{2}=v\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\frac{v}{a}-\frac{1}{2}a{\left(\frac{v}{a}\right)}^{2}\\ \Delta x=\frac{{v}^{2}}{2a}\end{array}$$
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