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The position of a particle related to time is given by x = ( 5 t 2 − 4 t + 5 ) m . The magnitude of velocity of the particle at t = 2 s will be :

Question

The position of a particle related to time is given by

x=(5t24t+5) m x = (5t^2 - 4t + 5) \text{ m}

The magnitude of velocity of the particle at
t = 2 s will be :

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Solution

To find the velocity of the particle at a given time, we need to take the derivative of the position function with respect to time. This is because velocity is the rate of change of position with respect to time.

The position function is given by x = 5t^2 - 4t + 5.

Taking the derivative of this function with respect to time gives us the velocity function:

v = dx/dt = d(5t^2 - 4t + 5)/dt = 10t - 4.

Now, to find the velocity at t = 2 s, we substitute t = 2 into the velocity function:

v = 10(2) - 4 = 20 - 4 = 16 m/s.

So, the magnitude of the velocity of the particle at t = 2 s is 16 m/s.

This problem has been solved

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