Given the position of an object s(t)=12t−t2s\left(t\right)=12t-t^2s(t)=12t−t2 (in meters), find the acceleration of the object at t=5t=5t=5 seconds.

Instantaneous Acceleration a = − 2 m s 2 a=-2\frac{m}{s^2} a = − 2 s 2 m

Given the position of an object s ( t ) = 12 t − t 2 s\left(t\right)=12t-t^2 s ( t ) = 12 t − t 2 (in meters), find the acceleration of the object at t = 5 t=5 t = 5 seconds.

Given the position of an object in meters described by this function right here, find the acceleration of the object at t equals 5 seconds. So in this problem, notice we're given a position function, and we're asked to find acceleration. Now the first step that I'm going to take is to take the derivative of position with respect to time, because I know the derivative of position is velocity. Now the derivative of 12 t, that t is just gonna come off, and we're going to be left with 12. Then that's gonna be minus the derivative of t squared, which using the power rule, that's going to come out to 2 t. So this right here is our is our function for velocity, but notice something. We're not asked for velocity in this problem. We're asked for acceleration. So to find the acceleration function, well, I need to take the derivative again, because acceleration is going to be the second derivative of our position or the first derivative of velocity. So if I take this derivative again, well, the derivative of 12, that's a constant, so we're just gonna get 0, And the derivative of negative two t is just gonna be negative 2. So that's gonna be our function for acceleration. So our acceleration function is going to be 0 minus 2, which we can also write as negative 2 meters per second squared. So this right here would be the result for acceleration, and notice something. Notice that we got that our acceleration is negative 2 meters per second squared constantly. What that means is that for every time t, no matter if it's 5 seconds, 10 seconds, or a 100 seconds, our acceleration is constantly going to be negative 2 meters per second squared. So that right there would be the solution to our problem, and that's how you can solve these types of problems where you're dealing with acceleration. Hope you found this video helpful.

Given the position of an object s ( t ) = 12 t − t 2 s\left(t\right)=12t-t^2 s ( t ) = 12 t − t 2 (in meters), find the acceleration of the object at t = 5 t=5 t = 5 seconds.

Instantaneous Acceleration a = − 2 m s 2 a=-2\frac{m}{s^2} a = − 2 s 2 m

Instantaneous Acceleration a = 9.8 m s 2 a=9.8\frac{m}{s^2} a = 9.8 s 2 m

Instantaneous Acceleration a = 2 m s 2 a=2\frac{m}{s^2} a = 2 s 2 m

Instantaneous Acceleration a = 0 m s 2 a=0\frac{m}{s^2} a = 0 s 2 m

4. Applications of Derivatives

Motion Analysis

Calculus

4. Applications of Derivatives

Motion Analysis

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Multiple Choice

Given the position of an object $s\left(t\right)=12t-t^2$ (in meters), find the acceleration of the object at $t=5$ seconds.

Instantaneous Acceleration $a=9.8\frac{m}{s^2}$

Instantaneous Acceleration $a=2\frac{m}{s^2}$

Instantaneous Acceleration $a=-2\frac{m}{s^2}$

Instantaneous Acceleration $a=0\frac{m}{s^2}$

Verified step by step guidance

To find the acceleration of the object, we first need to determine the velocity function by differentiating the position function s(t) with respect to time t.

The position function is given as s(t) = 12t - t^2. Differentiate this function to find the velocity function v(t).

The derivative of s(t) = 12t - t^2 with respect to t is v(t) = ds/dt = 12 - 2t.

Next, find the acceleration function by differentiating the velocity function v(t) with respect to time t.

The derivative of v(t) = 12 - 2t with respect to t is a(t) = dv/dt = -2. Evaluate this at t = 5 to find the acceleration at that specific time.

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