Initial velocity and acceleration of a p...

Initial velocity and acceleration of a particles are as shown in the figure. Acceleration vector of particle remain constant. Then radius of curvature of path of particle `:`

is 9 m initially

is `(9)/(sqrt(3))m` initially

will have minimum value of `(9)/(8) m`

will have minimum value `(3)/(8) m`

Text Solution

Verified by Experts

The correct Answer is:

A, C

v at origin `ne =`
`E=(r=2 m)=(K(-q)r)/((R_(1)^(2)+r^(2))^(3//2))+(K.Q.r)/((R_(2)^(2)+r^(2))^(3//2))`
`=K.rq[-(1)/(10^(3//2))+(2 sqrt(2))/(2^(3//2)10^(3//2))]=0`
From origin to r=2, field is towards origin.

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The figure shows th velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remain constant. The minimum radius of curvature of trajectory of the body is

`2m`

`3m`

`8m`

`16m`

Figure shows the velocity and acceleration of a point line body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is

2 m

3 m

8 m

16 m

The velocity displacement graph of a particle is shown in figure. The acceleration of the particle when x = 0 is.

`(1)/(sqrt(3))m//s^(2)`

`-(2)/(sqrt(3))m//s^(2)`

`(2)/(3)m//s^(2)`

None of the above