Correct Option is (A)

$|v_{10}|=(100+10\times 10)$ m/s

$v_{10}=-200$ m/s and $v_0=-100$ m/s

from 10s to 20s velocity remains zero

$\Rightarrow$ from t = 0s to 10s velocity increases in magnitude linearly.

$\Rightarrow$ graph given in option A fits correctly.

Correct Option is (A)

The slope of the given v versus x graph is $m=-\frac{v_0}{x_0}$ and intercept is c = + v₀. Hence, v varies with x as

$v=-\left(\frac{v_0}{x_0}\right)x+v_0$ ...... (1)

where v₀ and x₀ are constants of motion. Differentiating with respect to time t, we have

$\frac{dv}{dt}=-\left(\frac{v_0}{x_0}\right)\frac{dx}{dt}$

or $a=-\left(\frac{v_0}{x_0}\right)v$ ........ (2)

Using Eq. (1) in Eq. (2), we get

$a=-\left(\frac{v_0}{x_0}\right)\left(-\frac{v_0}{x_0}x+v_0\right)$

$a={\left(\frac{v_0}{x_0}\right)}^{2}x-\frac{v_0^2}{x_0}$

Thus the graph of a versus x is a straight line having a positive slope = ${\left(\frac{v_0}{x_0}\right)}^2$ and negative intercept = $-\frac{v_0^2}{x_0}$. Hence the correct choice is (a).

Correct Option is (B)

Acceleration(a) is constant

$\therefore$ v $\propto$ t (straight line graph)

and x $\propto$ t² (parabolic graph)

Correct Option is (C)

$t_1+t_2=t,$ $V^{\prime }=0+\alpha t_1$

$V=u+at$

$0=\alpha t_1-\beta t_2$

$t_2=\frac{\alpha }{\beta }{t}_1=t$

$t_1=\left(\frac{\beta }{\alpha +\beta }\right)t$

Distance $=\frac{1}{2}(t_1+t_2)\times \alpha t_1$ (area of triangle)

$=\frac{1}{2}t\times \alpha \left(\frac{\beta }{\alpha +\beta }\right)t$

$=\frac{\alpha \beta }{2(\alpha +\beta )}{t}^2$

Correct Option is (A)

We know that, $a=v\frac{dv}{dx}$

as slope is constant, so a $\propto$ v (from x = 0 to 200 m)

& slope = 0 so a = 0 (from x = 200 to 400 m)

$\therefore$