Question

59. The velocity (V) of a particle moving along x-axis
varies with its position x as shown in figure. The
acceleration (a) of particle varies with position (x) as
a= udv
(m/s)
x (m)
2
(1) a² = x+ 3
(3) 2a = 3x + 5
(2) a = 2x2 + 4
(4) a = 4x - 8​

Answer 1

Answer:

(4) a = 4x - 8

Explanation:

using slope intercept form, we find the equation for velocity in terms of position. Then we differentiate the equation with respect to time and substitute the value for velocity, to get the equation for acceleration

Answer 2

$\displaystyle\large\boxed {\sf {(4)\quad\!\!a=4x-8}}$

The particle attains $\displaystyle\sf {v=4}$ at $\displaystyle\sf {x=0}$ and $\displaystyle\sf {v=0}$ at $\displaystyle\sf {x=2.}$

Then the slope of the graph is,

$\displaystyle\longrightarrow\sf{\dfrac {dv}{dx}=\dfrac {4-0}{0-2}}$

$\displaystyle\longrightarrow\sf{\dfrac {dv}{dx}=\dfrac {4}{-2}}$

$\displaystyle\longrightarrow\sf{\dfrac {dv}{dx}=-2}$

$\displaystyle\longrightarrow\sf{dv=-2\ dx\quad\quad\dots (1)}$

Integrating (1),

$\displaystyle\longrightarrow\sf{\int dv=\int-2\ dx}$

$\displaystyle\longrightarrow\sf{v=-2x+c\quad\quad\dots (2)}$

But at $\displaystyle\sf {x=0,}$

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

From (2),

$\displaystyle\longrightarrow\sf{-2(0)+c=4}$

$\displaystyle\longrightarrow\sf{c=4}$

Then (2) becomes,

$\displaystyle\longrightarrow\sf{v=-2x+4 \quad\quad\dots(3)}$

Dividing each term in (1) by $\displaystyle\sf {dt,}$

$\displaystyle\longrightarrow\sf{\dfrac {dv}{dt}=-2\cdot\dfrac {dx}{dt}}$

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

From (3),

$\displaystyle\longrightarrow\sf{a=-2(-2x+4)}$

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

$\displaystyle\longrightarrow\sf {\underline {\underline {a=4x-8}}}$