A particle moves along the x-axis so that at time t = 0 its position is given by x(t)=4t^(3)-36t^(2). Determine all intervals when the speed of the particle is increasing.

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A particle moves along the  x-axis so that at time  t >= 0 its position is given by  x(t)=4t^(3)-36t^(2). Determine all intervals when the speed of the particle is increasing.

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Find Velocity Function: To determine when the speed of the particle is increasing, we first need to find the velocity of the particle, which is the derivative of the position function x(t). The position function is given by x(t) = 4t^3 - 36t^2. Let's find the derivative of x(t) with respect to t to get the velocity function v(t). v(t) = dx(t)/dt = d/dt (4t^3 - 36t^2) = 12t^2 - 72t.Find Acceleration Function: Next, we need to find the acceleration of the particle, which is the derivative of the velocity function v(t). a(t) = dv(t)/dt = d/dt (12t^2 - 72t) = 24t - 72.Determine Sign of Acceleration: The speed of the particle is increasing when the velocity and acceleration have the same sign, that is, when both are positive or both are negative. To find when this occurs, we need to determine the sign of the acceleration function a(t) = 24t - 72. We set the acceleration equal to zero to find the critical points: 24t - 72 = 0. Solving for t gives us t = 72/24 = 3.Test Intervals Around Critical Point: We now have a critical point at t = 3. We need to test the intervals around this point to determine the sign of the acceleration. Let's choose test points in the intervals (0, 3) and (3, infinity). For t = 2 (in the interval (0, 3)), a(2) = 24(2) - 72 = 48 - 72 = -24, which is negative. For t = 4 (in the interval (3, infinity)), a(4) = 24(4) - 72 = 96 - 72 = 24, which is positive.Check Velocity Sign at Critical Point: Since the acceleration is negative when t < 3 and positive when t > 3, the speed of the particle is increasing for t > 3. We also need to check the sign of the velocity at t = 3 to ensure that it is not zero, as the speed would not be increasing at a point where the velocity is zero. v(3) = 12(3)^2 - 72(3) = 108 - 216 = -108, which is negative.Conclusion: Given that the velocity at t = 3 is negative and the acceleration is positive for t > 3, the speed of the particle is indeed increasing for t > 3. Therefore, the interval when the speed of the particle is increasing is (3, infinity).

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=4 t^{3}-36 t^{2}$ . Determine all intervals when the speed of the particle is increasing.

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A particle moves along the x x x-axis so that at time t≥0 t \geq 0 t≥0 its position is given by x(t)=4t3−36t2 x(t)=4 t^{3}-36 t^{2} x(t)=4t3−36t2. Determine all intervals when the speed of the particle is increasing.

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=4 t^{3}-36 t^{2}$ . Determine all intervals when the speed of the particle is increasing.