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At time t, the position of a body moving along the s-axis is s=t3−6t2+9tm. a. Find the body's acceleration each time the velocity is zero. b. Find the body's speed each time the acceleration is zero. c. Find the total distance traveled by the body from t=0 to t=2.

User Flurbius
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Final answer:

a. When the velocity is zero, the body's acceleration is 0 m/s^2. b. When the acceleration is zero, the body's speed is 9 m/s. c. The total distance traveled by the body from t = 0 to t = 2 is 10 m.

Step-by-step explanation:

a. To find the body's acceleration when the velocity is zero, we need to find the values of t when the velocity is zero. The velocity of the body is given by v(t) = s'(t), which is the derivative of the position function with respect to time. The acceleration of the body is given by a(t) = v'(t), which is the derivative of the velocity function with respect to time. So, first, we find the velocity function as v(t) = s'(t) = 3t^2 - 12t + 9

To find when the velocity is zero, we set v(t) = 0 and solve for t:

0 = 3t^2 - 12t + 9

By factoring: 0 = (t-1)(3t-9)

This gives us t = 1 and t = 3. So, when the velocity is zero, the body's acceleration is a(1) = 3(1)^2 - 12(1) + 9 = 0 m/s^2 and a(3) = 3(3)^2 - 12(3) + 9 = 0 m/s^2.

b. To find the body's speed when the acceleration is zero, we need to find the values of t when the acceleration is zero. The acceleration of the body is given by a(t) = v'(t), which is the derivative of the velocity function with respect to time. So, to find when the acceleration is zero, we need to find when v'(t) = 0. The acceleration function is a(t) = v'(t) = 6t - 12.

Setting a(t) = 0, we have 6t - 12 = 0. Solving for t gives t = 2. So, when the acceleration is zero, the body's speed is given by the magnitude of the velocity function at t = 2: |v(2)| = |3(2)^2 - 12(2) + 9| = 9 m/s.

c. To find the total distance traveled by the body from t = 0 to t = 2, we need to find the definite integral of the speed function from t = 0 to t = 2. The speed function is given by |v(t)| = |3t^2 - 12t + 9|.

Integrating the speed function over the interval [0,2]:

D = ∫[0,2] |v(t)| dt = ∫[0,2] (3t^2 - 12t + 9) dt = [(t^3 - 6t^2 + 9t)]_[0,2] = (2^3 - 6(2)^2 + 9(2)) - (0^3 - 6(0)^2 + 9(0)) = 10 m.

User Ben Walker
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