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Math

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PROBLEM

Find the car's acceleration at t=2t=2 seconds, given s(t)=2t49t36t28s(t)=2 t^{4}-9 t^{3}-6 t^{2}-8.

STEP 1

Assumptions1. The position of the car is given by the function s(t)=t49t36t8s(t)= t^{4}-9 t^{3}-6 t^{}-8
. The variable tt is measured in seconds3. The variable ss is measured in meters4. We are looking for the acceleration of the car at t=t= seconds

STEP 2

The acceleration of the car is the second derivative of the position function. So first, we need to find the first derivative of the position function, which will give us the velocity function.
v(t)=s(t)v(t) = s'(t)

STEP 3

Now, we calculate the first derivative of the position function.
v(t)=ddt(2t9t36t28)v(t) = \frac{d}{dt}(2 t^{}-9 t^{3}-6 t^{2}-8)

STEP 4

Using the power rule for differentiation, we getv(t)=8t327t212tv(t) =8t^{3} -27t^{2} -12t

STEP 5

Now that we have the velocity function, we can find the acceleration function by taking the derivative of the velocity function.
a(t)=v(t)a(t) = v'(t)

STEP 6

Now, we calculate the second derivative of the position function, which is the first derivative of the velocity function.
a(t)=ddt(8t327t212t)a(t) = \frac{d}{dt}(8t^{3} -27t^{2} -12t)

STEP 7

Using the power rule for differentiation again, we geta(t)=24t254t12a(t) =24t^{2} -54t -12

STEP 8

Now that we have the acceleration function, we can find the acceleration of the car at t=2t=2 seconds by substituting t=2t=2 into the acceleration function.
a(2)=24(2)254(2)12a(2) =24(2)^{2} -54(2) -12

SOLUTION

Calculate the acceleration at t=2t=2 seconds.
a(2)=24(4)54(2)12=9610812=24a(2) =24(4) -54(2) -12 =96 -108 -12 = -24The acceleration of the car at t=2t=2 seconds is 24-24 m/s².

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