The velocity \(v ms^{-1}\) of a particle moving in a straight line is given by \(v = 3t^{2} - 2t + 1\) at time t secs. Find the acceleration of the particle...

The velocity \(v ms^{-1}\) of a particle moving in a straight line is given by \(v = 3t^{2} - 2t + 1\) at time t secs. Find the acceleration of the particle after 3 seconds.

Answer Details

To find the acceleration of the particle, we need to take the derivative of the velocity function with respect to time (t). The derivative of velocity with respect to time gives us the acceleration. Taking the derivative of v with respect to t, we get: \[\frac{dv}{dt} = 6t - 2\] Now, we need to find the acceleration of the particle after 3 seconds. We can do this by substituting t = 3 into the expression we just found for the derivative of v: \[\frac{dv}{dt} = 6t - 2\] \[\frac{dv}{dt}\bigg|_{t=3} = 6(3) - 2\] \[\frac{dv}{dt}\bigg|_{t=3} = 16\] Therefore, the acceleration of the particle after 3 seconds is 16 \(ms^{-2}\). To summarize, we found the derivative of the velocity function with respect to time to get the acceleration function. Then, we substituted t = 3 into the acceleration function to find the acceleration of the particle after 3 seconds.