Question

a particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. the particle's position at t0=0s is x0 = -5.40 m . at t1 = 2.00 s , the particle is at x1 = 5.80 m .

Answer 1

The acceleration of the particle is -0.25t^(-2) s^(-2). The position of the particle at t = 2.0 s is 5.80 m.

Step-by-step explanation:

The given velocity function is v(t) = A + Bt^(-1) with A = 2 m/s and B = 0.25 m. To determine the acceleration, we need to take the derivative of the velocity function with respect to time, which gives us a(t) = -Bt^(-2). In this case, a(t) = -0.25t^(-2) s^(-2).

To find the position of the particle at t = 2.0 s, we need to integrate the velocity function over the interval [0, 2.0]. The position function is given by x(t) = ∫[0, t]v(t) dt + x0, where x0 = -5.40 m is the initial position. Plugging in the values, we have x(2.0) = ∫[0, 2.0]kt^2 dt - 5.40. Evaluating the integral gives us x(2.0) = (k/3)(2.0^3) - 5.40 = 5.80 m, which matches the given value.