Question

Please answer this quickly​

Answer 1

We see the graph passes through the points (0, 10) and (5, 0).

The slope of the graph,

$\sf{\longrightarrow m=\dfrac{10-0}{0-5}}$

$\sf{\longrightarrow m=-2}$

By slope - intercept form, the equation of the graph will be,

$\sf{\longrightarrow v=-2x+10}$

Differentiating wrt time $\sf{t,}$

$\sf{\longrightarrow \dfrac{dv}{dt}=\dfrac{d}{dt}[-2x+10]}$

$\sf{\longrightarrow a=\dfrac{d}{dx}[-2x+10]\cdot\dfrac{dx}{dt}}$

$\sf{\longrightarrow a=-2v}$

$\sf{\longrightarrow a=-2(-2x+10)}$

$\sf{\longrightarrow a=4x-20}$

Again differentiating wrt time $\sf{t,}$

$\sf{\longrightarrow \dfrac{da}{dt}=\dfrac{d}{dt}[4x-20]}$

Let $\sf{\dfrac{da}{dt}=j}$ whose SI unit is $\sf{m\,s^{-3}.}$

$\sf{\longrightarrow j=\dfrac{d}{dx}[4x-20]\cdot\dfrac{dx}{dt}}$

$\sf{\longrightarrow j=4v}$

$\sf{\longrightarrow j=4(-2x+10)}$

$\sf{\longrightarrow j=-8x+40}$

At $\sf{x=2\ m,}$

$\sf{\longrightarrow a(2)=4(2)-20}$

$\sf{\longrightarrow a(2)=-12\ m\,s^{-2}}$

At $\sf{x=3\ m,}$

$\sf{\longrightarrow a(3)=4(3)-20}$

$\sf{\longrightarrow a(3)=-8\ m\,s^{-2}}$

and,

$\sf{\longrightarrow j(3)=-8(3)+40}$

$\sf{\longrightarrow j(3)=16\ m\,s^{-3}}$

At $\sf{x=4\ m,}$

$\sf{\longrightarrow j(4)=-8(4)+40}$

$\sf{\longrightarrow j(4)=8\ m\,s^{-3}}$

Hence 1st and 2nd options are correct.

Answer 2

Step-by-step explanation:

1st and 2nd options are correct.