The acceleration a(t) of a moving object is given by the derivative of the velocity function v(t). Find the acceleration of the object at time t=5 seconds if v(t)=12tsquared+8t meters per second. Thanks for any help.
v = 12 t^2 + 8 t
dv/dt = a = 24 t + 8
a(5) = 24*5 + 8
= 128
To find the acceleration of the object at time t=5 seconds, we need to take the derivative of the velocity function v(t) with respect to time.
Given the velocity function v(t) = 12t^2 + 8t meters per second, we can find the acceleration function a(t) by taking the derivative of v(t) with respect to t.
To do this, we can apply the power rule and sum rule of differentiation.
The power rule states that if we have a function f(t) = kt^n, then the derivative is given by f'(t) = nk*t^(n-1), where k is a constant.
In our case, v(t) = 12t^2 + 8t, and applying the power rule, we get v'(t) = d/dt(12t^2) + d/dt(8t).
Differentiating each term separately, we get:
v'(t) = 24t + 8.
Therefore, the acceleration function a(t) is given by a(t) = v'(t) = 24t + 8.
Now, to find the acceleration at time t = 5 seconds, we substitute t = 5 into the acceleration function:
a(5) = 24(5) + 8
= 120 + 8
= 128 meters per second squared.
So, the acceleration of the object at time t = 5 seconds is 128 meters per second squared.
I hope this explanation helps! Let me know if you have any further questions.