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. We differentiate x(t) to get the velocity function v(t). v(t) = dx(t)/dt = d/dt (-5t^4 + 30t^2) = -20t^3 + 60t.Find 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. The velocity function is v(t) = -20t^3 + 60t. We differentiate v(t) to get the acceleration function a(t). a(t) = dv(t)/dt = d/dt (-20t^3 + 60t) = -60t^2 + 60.Determine Speed Increase: To determine when the speed of the particle is increasing, we need to find when the velocity and acceleration have the same sign, as speed is the magnitude of velocity and increases when velocity is being ＂pushed＂ by acceleration in the same direction. We set up inequalities to find when v(t) and a(t) are both positive or both negative.Find Positive Velocity: First, let's find when v(t) is positive. v(t) = -20t^3 + 60t > 0. We factor out 20t from v(t) to solve the inequality. 20t(-t^2 + 3) > 0. The critical points are t = 0 and t = sqrt(3). We create a sign chart to determine the intervals where v(t) is positive.Find Positive Acceleration: For the sign chart, we test values in the intervals (-∞, 0), (0, sqrt(3)), and (sqrt(3), ∞). For t < 0, v(t) is positive (since -t^2 + 3 is always positive when t is real and t^2 is less than 3). For 0 < t < sqrt(3), v(t) is positive. For t > sqrt(3), v(t) is negative (since -t^2 + 3 becomes negative). So, v(t) is positive in the intervals (-∞, 0) and (0, sqrt(3)).Combine Positive Intervals: Next, let's find when a(t) is positive. a(t) = -60t^2 + 60 > 0. We factor out 60 from a(t) to solve the inequality. 60(-t^2 + 1) > 0. The critical point is t = 1. We create a sign chart to determine the intervals where a(t) is positive.For the sign chart, we test values in the intervals (-∞, 1) and (1, ∞). For t < 1, a(t) is positive (since -t^2 + 1 is positive when t^2 is less than 1). For t > 1, a(t) is negative (since -t^2 + 1 becomes negative). So, a(t) is positive in the interval (-∞, 1).Now we combine the intervals where both v(t) and a(t) are positive to find when the speed is increasing. The intervals where v(t) is positive are (-∞, 0) and (0, sqrt(3)). The interval where a(t) is positive is (-∞, 1). The only interval where both v(t) and a(t) are positive is (0, 1). Therefore, 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.