Question

the position of a particle moving on a straight track is given by {x=(6+9t+3t^2)m} what is acceleration 'a' at t=2 second?

Answer 1

Answer:

sorry I don't know the answer

Answer 2

Answer: Acceleration 'a' at t=2 second is equal to 6 m/s²

Explanation:

x =  6 + 9t + 3t²  m

Velocity is defined as rate of change of position with respect to time.

In given question, body is moving in straight line along x axis. Hence, only x coordinate of the body is changing, which will give velocity only in x-direction.

On differentiating position with respect to time, we get velocity.

Therefore,

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

$v = \frac{d(3t^{2}+9t +6) }{dt}$

v =  6t + 9  m/s

Acceleration of a body is defined as the rate of change of velocity with respect to time. On differentiating velocity with respect to time, we get acceleration.

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

$a = \frac{d(6t+9)}{dt}$

a = 6 m/s²

Acceleration of the body is independent of time. At any given time t, the value of acceleration is 6 m/s².

Therefore,

At t = 2 seconds, a = 6 m/s².

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