Physics Motion in Straight Line

Suppose the position-time graph is a straight line, in this case, the velocity is constant. This means that there is no acceleration.
If the graph is curved, velocity is changing, which means that there is acceleration. If the graph is concave, the slopes will get more positive with time. This means that there is positive acceleration. If the graph is cap-shaped, the slope will become more negative with time. This is known as negative acceleration.

To graph motion in a straight line, you need to visualise the relationship between different kinematic quantities like position, velocity and time. Suppose an object moves with a constant velocity, the position-time graph will be a straight line with constant slope. If the object accelerates, the slope of position-time graph will change with time and result in a curved line.

Sign conventions are important because all kinematic variables can be positive or negative. You must first choose your coordinate system and positive direction, then consistently apply signs. For example, if you are choosing upward as positive in free-fall problems, gravity becomes negative (a = -g), upward initial velocity is positive, and downward displacement is negative. The equations of motion work for any situation, as long as you substitute values with proper signs. Incorrect conventions lead to wrong answers.

It's not always true. Direction of velocity is the most important consideration here that will tell us whether acceleration increases or decreases speed. You can consider two scenarios. If you're falling, that's negative velocity. That implies negative gravitational acceleration. Here, your speed increases. If you're moving upward, that's positive velocity. Now, with that same negative acceleration, your speed decreases. For choosing the equations of motion, you need to know both the signs of acceleration and velocity to determine if you're speeding up or slowing down.

Zero velocity doesn't mean zero acceleration. When you throw a ball-like object upward, at its peak the velocity is zero. But acceleration remains constant due to gravity. The velocity is still changing from positive to negative at that instant. That means acceleration continues.

Let the distance travelled is x, so

$$ $\left|{\overrightarrow{v}}_{Eg}\right|=\frac{x}{t_2}\Rightarrow$ speed of escalator for ground

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{BE}\right|=\frac{x}{t_1}\Rightarrow$ speed of boy concerning the escalator

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{Bg}\right|=\frac{x}{t_1}+\frac{x}{t_2}\Rightarrow$ speed of boy concerning ground

The time taken by him to walk up the moving escalator = t =  $\frac{x}{\left|{\overrightarrow{v}}_{Bg}\right|}$ $\frac{}{{\stackrel{}{}}_{}}$

$\Rightarrow t=\frac{x}{\frac{x}{t_1}+\frac{x}{t_2}}=\frac{t_1{t}_2}{t_1+t_2}$

Equation from the given graph:

$\begin{array}{l}{v}^2=2x+20\\ v^2=2x+20\\ \Rightarrow 2V\frac{dV}{dt}=2V\end{array}$

=>a = 1m/s²

Time taken by first drop to reach the ground,

$t_1=\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2*9.8}{9.8}}=\sqrt[]{2}s$              

Time for which second drop has fallen,

$t_2=t_1-\frac{t_1}{2}=\sqrt[]{2}-\frac{1}{\sqrt[]{2}}s$

Height of the second drop

$=9.8-\frac{1}{2}*9.8*{\left(\sqrt[]{2}-\frac{1}{\sqrt[]{2}}\right)}^2=7.35m$

$a=V\frac{dv}{dx}$

$\Rightarrow a=\sqrt[]{5000+24x}*\frac{12}{\sqrt[]{5000+24x}}$

=> a = 12 m/s²