A particle moves along the x-axis so that at time t = 0 its position is given by x(t)=-5t^(4)+30t^(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)=-5t^(4)+30t^(2). Determine all intervals when the speed of the particle is increasing.

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Find Velocity Function: First, we need to find the velocity of the particle, which is the derivative of the position function x(t) with respect to time t. The position function is x(t) = -5t^4 + 30t^2. Let's find the derivative of x(t) to get the velocity function v(t). v(t) = dx/dt = d/dt (-5t^4 + 30t^2) v(t) = -20t^3 + 60tFind Acceleration Function: Next, we need to find the acceleration of the particle, which is the derivative of the velocity function v(t) with respect to time t. Let's find the derivative of v(t) to get the acceleration function a(t). a(t) = dv/dt = d/dt (-20t^3 + 60t) a(t) = -60t^2 + 60Determine Critical Points: The speed of the particle is increasing when the velocity and acceleration have the same sign, either both positive or both negative. To find when they have the same sign, we need to determine the sign of the acceleration function a(t) since the velocity function v(t) can change signs. Let's find the critical points of the acceleration function by setting a(t) equal to zero and solving for t. -60t^2 + 60 = 0 t^2 = 1 t = ±1Test Acceleration Intervals: We have two critical points, t = -1 and t = 1. However, since the problem states that t >= 0, we only consider t = 1. Now we need to test the intervals around t = 1 to determine where the acceleration is positive (since t = -1 is not in our domain). We choose test points in the intervals (0, 1) and (1, ∞) and plug them into the acceleration function a(t). Let's choose t = 0.5 for the interval (0, 1) and t = 2 for the interval (1, ∞). a(0.5) = -60(0.5)^2 + 60 = -60(0.25) + 60 = -15 + 60 = 45 a(2) = -60(2)^2 + 60 = -60(4) + 60 = -240 + 60 = -180Check Velocity Sign: Since a(0.5) is positive, the acceleration is positive in the interval (0, 1), and since a(2) is negative, the acceleration is negative in the interval (1, ∞). Now we need to check the sign of the velocity function v(t) in the interval (0, 1) to ensure that the speed is increasing. Let's choose the same test point t = 0.5 and plug it into the velocity function v(t). v(0.5) = -20(0.5)^3 + 60(0.5) = -20(0.125) + 30 = -2.5 + 30 = 27.5 Since v(0.5) is positive and a(0.5) is also positive, the speed of the particle is increasing in the interval (0, 1).

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=-5 t^{4}+30 t^{2}$ . 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)=−5t4+30t2 x(t)=-5 t^{4}+30 t^{2} x(t)=−5t4+30t2. 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 $x(t)=-5 t^{4}+30 t^{2}$ . Determine all intervals when the speed of the particle is increasing.