The velocity of a particle is v=(2t^2-4t 3) m/s where t is tha time in second .find tha acceleration of this particle at t=2 seconds.?

Question

Answer

A=(4*2)-4= 4m/s^2

Answer

Given:

The velocity of a particle is given by the equation v = 2t^2 - 4t + 3 m/s, where t is the time in seconds.

To Find:

The acceleration of the particle at t = 2 seconds.

Explanation:

To find the acceleration, we need to differentiate the velocity function with respect to time. The derivative of velocity with respect to time gives us the acceleration.

Given, v = 2t^2 - 4t + 3 m/s

Differentiating both sides of the equation with respect to time (t), we get:

dv/dt = d/dt (2t^2 - 4t + 3)

Derivation:

Using the power rule of differentiation, we differentiate each term separately:

d/dt (2t^2) = 2 * 2t^1 = 4t
d/dt (-4t) = -4
d/dt (3) = 0

Therefore, dv/dt = 4t - 4

Substituting t = 2:

To find the acceleration at t = 2 seconds, we substitute t = 2 into the equation dv/dt = 4t - 4:

dv/dt = 4t - 4
dv/dt = 4(2) - 4
dv/dt = 8 - 4
dv/dt = 4 m/s^2

Therefore, the acceleration of the particle at t = 2 seconds is 4 m/s^2.

Conclusion:

The acceleration of the particle at t = 2 seconds is 4 m/s^2. This means that the rate of change of velocity with respect to time at t = 2 is 4 m/s^2.

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