Question

37. Position of a particle varies with time t(ins) as r = (2ti^ + 3tj^)m. The magnitude of acceleration of particle is
(1) 2 m/s2
(2) 4 m/s2
(3) 5 m/s2
(4) Zero​

Answer 1

Answer:

(D) Zero

Explanation:

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Answer 2

The magnitude of acceleration of particle is (4)Zero.

Given,

Equation of the position of the particle that is varying with time, r=(2ti + 3tj) m.

To find,

the magnitude of acceleration of particle.

Solution:

• Velocity is equal to the rate of displacement of an object or the rate of change of positions.
• It is represented by the following expression:
• $v=\frac{ds}{dt}$ or $v=\frac{dr}{dt}$.
• where, ds-displacement and dv-position.
• Acceleration of an object is equal to the rate of change of velocity.
• It is represented by the following expression:
• $a=\frac{dv}{dt}$ .

The velocity of the particle will be:

$v=\frac{dr}{dt}\\v=\frac{d(2ti+3tj)}{dt} \\v=(2i+3j) m/sec.$

The acceleration of the particle will be:

$a=\frac{dv}{dt} \\a=\frac{d(2i+3j)}{dt} \\a=0.$

As the equation of acceleration comes as zero its magnitude will also be zero.

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