Question

the particle moves along X axis as X = 4(t-2) + a(t-2)^2. which of the following is true? a) the initial velocity of particle is 4. b) the acceleration of particle is 2a. c) the particle is at origin at t = 0. ​

Answer 1

Answer:

any doubt msg me,I explained above in pic

Answer 2

None of the option is true.

Explanation:

It is given that,

The particle moves along x axis as given by the following equation as :

$x=4(t-2)+a(t-2)^2$

We know that the velocity of the particle is given by :

$v=\dfrac{dx}{dt}$

$v=\dfrac{d(4(t-2)+a(t-2)^2)}{dt}$

$v=4+2a(t-2)$

Initial velocity of the particle is the velocity when t = 0 s

$v=(4-4a)\ m/s$

Option (1) is incorrect.

We know that the acceleration of the particle is given by :

$a=\dfrac{dv}{dt}$

$a=\dfrac{d(4+2a(t-2))}{dt}$

$a=2a\ m/s^2$

Option (2) is incorrect.

At t = 0,

The position of particle is :

$x=4(0-2)+a(0-2)^2$

$x=(-8+4a)\ m$

Particle is not at origin at t = 0, it depends on the value of a.

Hence, this is the required solution.

Learn more,

Kinematics

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