The following question refers to the following expression for a particle's position:
X = 3t^2 - 4t + 3
What is the particle's acceleration after 3 seconds?
I keep getting 1 m/s2 but that's not an answer choice.
I found instantaneous velocity at t=0 (3 m/s) and at t = 3 (6 m/s) and just calculated change in velocity over change in time....
Where am I going wrong?
To find the particle's acceleration after 3 seconds, you need to take the second derivative of the position function with respect to time (t).
Given that the position function is X = 3t^2 - 4t + 3, let's begin by calculating the first derivative, which will give us the velocity function:
V = dX/dt = d(3t^2 - 4t + 3)/dt
To differentiate the function, use the power rule of differentiation.
V = 6t - 4
Now, to find the acceleration, we need to take the derivative of the velocity function:
A = dV/dt = d(6t - 4)/dt
Again, using the power rule, we find:
A = 6
Therefore, the particle's acceleration after 3 seconds is 6 m/s^2.
It seems that you made a calculation error while finding the instantaneous velocity at t = 3. It should be V = 6t - 4, so at t = 3, the velocity should be V = 6(3) - 4 = 14 m/s, not 6 m/s.
To calculate the acceleration, you should find the change in velocity over the change in time. In this case, the change in velocity is V - V₀ where V is the final velocity and V₀ is the initial velocity. In your case, V = 14 m/s and V₀ = 3 m/s. And the change in time is 3 seconds.
So, the acceleration = (V - V₀) / Δt = (14 - 3) / 3 = 11 / 3 = 3.67 m/s².
Therefore, the correct answer for the particle's acceleration after 3 seconds is approximately 3.67 m/s².