A particle is moving with constant acceleration 'a'. Following graph shows v^{2} versus x (d

English

Gujarati

Hindi

2.Motion in Straight Line

hard

A particle is moving with constant acceleration $'a'.$ Following graph shows $v^{2}$ versus $x$ (displacement) plot. The acceleration of the particle is $......{m} / {s}^{2}$

A$100$

B$20$

C$14$

D$1$

(JEE MAIN-2021)

Solution

${y}={m} {x}+{C}$
${v}^{2}=\frac{20}{10} {x}+20$
${v}^{2}=2 {x}+20$
$2 {v} \frac{{d} {v}}{{d} {x}}=2$
$\therefore {a}={v} \frac{{d} {v}}{{d} {x}}=1$

Standard 11

Physics

Share

Similar Questions

A car is moving along a straight road with a uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocity $30\; km / h$ and $40\; km / h$ respectively. The velocity of the car midway between $P$ and $Q$

medium

(AIPMT-1988)

Particle $A$ is moving along $x$-axis. At time $t=0$, it has velocity of $10\,m / s$ and acceleration $-4\,m / s ^2$. Particle $B$ has velocity of $20\,m / s$ and acceleration $-2\,m / s ^2$. Initially, both the particles are at origin. At time $t=2\,s$, distance between the two particles is $………….\,m$

medium

Brakes are being applied in a car moving with speed $20\ m/s$ which produces a retardation of $5\ m/s^2$ . The distance travelled by car till it stops is……..$m$

easy

For a body moving with uniform acceleration along straight line, the variation of its velocity $(v)$ with position $(x)$ is best represented by

easy

A particle starts moving from rest in a straight line with constant acceleration. After time $t_0$, acceleration changes its sign (just opposite to the initial direction), remaining the same in magnitude. Determine the time from the beginning of motion in which the particle returns to the initial position.

hard