Correct Option is (C)

Given that,

acceleration (a) = $-$ C (constant)

$\therefore$ $\frac{dv}{dt}$ = $-$ c

$\Rightarrow$$$ $\frac{dv}{dx}$ . $\frac{dx}{dt}$ = $-$ c

$\Rightarrow$$$ $\upsilon$ $\frac{dv}{dx}$ = $-$ c

$\Rightarrow$$$ $\int vdv$ = $-$ c$\int dx$

$\Rightarrow$$$ $\frac{v^2}{2}$ = $-$ cx + k

$\Rightarrow$$$ x = $-$ $\frac{v^2}{2c}$ + $\frac{k}{c}$

From this equation we can say option (c) is the correct graph.

Correct Option is (D)

Motion of the particle is shown below



Initially speed of the particle is maximum(v₀) and at heigth h velocity will become zero, then particle will move downward and velocity will increase gradually and become maximum when it reaches the ground. And acceleration in this entire motion will be -g, so slope of v - t will be same and negative.

So only Option (D) will be correct.

Correct Option is (A)

Using $h=ut+\frac{1}{2}g{t}^2$

$y_1=10t-5t^2;y_2=40t-5t^2$

$\therefore$ $y_2-y_1=30t$ for $t\le 8s.$

Curve will be straight line when $t\le 8s.$

when stone 1 reaches the ground then $y_1=-240m,andt=8s$

for $t>8s.$

$y_2-y_1=40t+\frac{1}{2}g{t}^2-240$

So, it will be a parabolic curve till stone 2 reaches the ground. And parabola should opens upward as coefficient of t² is positive.

Correct Option is (B)

For downward motion :

$v=-gt$

The velocity of the rubber ball increases in downward direction and we get a straight line between $v$ and $t$ with a negative slope.

Also applying $y-y_0=ut+\frac{1}{2}a{t}^2$

We get $y-h=-\frac{1}{2}g{t}^2\Rightarrow y=h-\frac{1}{2}g{t}^2$

The graph between $y$ and $t$ is a parabola with $y=h$ at $t=0.$ As time increases $y$ decreases.

For upward motion :
The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here $v=u-gt$ where $u$ is the velocity just after collision. As $t$ increases, $v$ decreases. We get a straight line between $v$ and $t$ with negative slope.

Also $y=ut-\frac{1}{2}g{t}^2$

All these characteristics are represented by graph $(B).$

Correct Option is (B)

For the body starting from rest

$x_1=0+\frac{1}{2}a{t}^2\Rightarrow x_1=\frac{1}{2}a{t}^2$

For the body moving with constant speed

$x_2=vt$

$\therefore$ $x_1-x_2=\frac{1}{2}a{t}^2-vt$

$\Rightarrow \frac{d\left(x_1-x_2\right)}{dt}=at-v$

at $t=0,$ $x_1-x_2=0$, so graph should start from origin.

For $at<v;$ the slope is negative that means $x_1-x_2$ < 0 so initially velocity of 1st body is less than second body and velocity of 1st body is increasing gradually.

For $at=v;$ the slope is zero. So $x_1-x_2$ = 0 it means here velocity of both the bodies are same.

For $at>v;$ the slope is positive. So $x_1-x_2$ > 0 it means here velocity of first body is greater than second body.

We know the relation between distance and time is.

$S=ut+\frac{1}{2}a{t}^2$, which is a equation parabola. So the graph should be a parabola.

These characteristics are represented by graph $(b).$