Motion in a Straight Line

Please find the solution below:

[h] = ML²T^-1

[E] = ML²T^-2

[V] = ML²T^-2C^-1

[P] = MLT^-1

According to question, we can write

 10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$                        

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$  

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$  

Average speed $=\left(\frac{4{\mathrm{v}}^2}{3\mathrm{v}}\right)$ $\left(\frac{{}^{}}{}\right)$

$=\frac{4\mathrm{v}}{3}$

(d) Initial velocity  $=-v\hat{j}$ $\stackrel{}{}$

Final velocity $=v\hat{i}$ $\stackrel{}{}$

Change in velocity  $=v\hat{i}-(-v\hat{j})$

$\stackrel{}{}\stackrel{}{}$$\stackrel{}{}\stackrel{}{}$ $=v(\hat{i}+\hat{j})$ Momentum gain is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$ $\stackrel{}{}\stackrel{}{}$

$$ $\Rightarrow$  Force experienced is along $\hat{i}+\hat{j}$ $\stackrel{}{}\stackrel{}{}$

$$  $\Rightarrow$ Force experienced is in North-East direction.

Air resistance resists the motion of an object. In this case, the net acceleration is lesser than 'g' and it shrinks as the speed increases. This makes the object to speed up more slowly. Ultimately, it reaches a constant terminal velocity which is lower for large-area ones and higher for heavy and streamlined ones.

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.

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 =  $\frac{x}{\left|{\overrightarrow{v}}_{Bg}\right|}$ $\frac{}{{\stackrel{}{}}_{}}$

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