A particle moves along the x-axis so that at time t = 0 its position is given by x(t)=t^(3)+5t^(2)-32 t. Determine if the particle is speeding up or slowing down at t=1.

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A particle moves along the  x-axis so that at time  t >= 0 its position is given by  x(t)=t^(3)+5t^(2)-32 t. Determine if the particle is speeding up or slowing down at  t=1.

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Find Velocity Function: To determine if the particle is speeding up or slowing down at t=1, we need to find the velocity and acceleration at that point in time. The velocity is the first derivative of the position function with respect to time, and the acceleration is the second derivative of the position function with respect to time.Evaluate Velocity at t=1: First, let's find the velocity function, v(t), by differentiating the position function x(t) with respect to t. v(t) = dx(t)/dt = d(t^3 + 5t^2 - 32t)/dt v(t) = 3t^2 + 10t - 32 Now we can evaluate the velocity at t=1. v(1) = 3(1)^2 + 10(1) - 32 v(1) = 3 + 10 - 32 v(1) = 13 - 32 v(1) = -19Find Acceleration Function: Next, we find the acceleration function, a(t), by differentiating the velocity function v(t) with respect to t. a(t) = dv(t)/dt = d(3t^2 + 10t - 32)/dt a(t) = 6t + 10 Now we can evaluate the acceleration at t=1. a(1) = 6(1) + 10 a(1) = 6 + 10 a(1) = 16Evaluate Acceleration at t=1: To determine if the particle is speeding up or slowing down at t=1, we look at the signs of the velocity and acceleration. If both the velocity and acceleration have the same sign (both positive or both negative), the particle is speeding up. If they have opposite signs, the particle is slowing down. At t=1, we found that v(1) = -19 (negative) and a(1) = 16 (positive), which means they have opposite signs.Determine Speeding Up or Slowing Down: Since the velocity and acceleration have opposite signs at t=1, the particle is slowing down at that moment.

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=t^{3}+5 t^{2}-32 t$ . Determine if the particle is speeding up or slowing down at $t=1$ .

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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)=t3+5t2−32t x(t)=t^{3}+5 t^{2}-32 t x(t)=t3+5t2−32t. Determine if the particle is speeding up or slowing down at t=1 t=1 t=1.

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=t^{3}+5 t^{2}-32 t$ . Determine if the particle is speeding up or slowing down at $t=1$ .