Question

The velocity of a particle moving along the x-axis varies with time according to v(t)=A+Bt−1, where A = 2 m/s, B = 0.25 m, and 1.0s≤t≤8.0s. Determine the acceleration and position of the particle at t = 2.0 s and t = 5.0 s. Assume that x(t=1s)=0.

a) Option a: a(2.0 s) = -0.25 m/s^2, x(2.0 s) = 3.25 m; a(5.0 s) = -0.25 m/s^2, x(5.0 s) = 5.875 m
b) Option a: a(2.0 s) = 0.25 m/s^2, x(2.0 s) = 2.75 m; a(5.0 s) = 0.25 m/s^2, x(5.0 s) = 7.875 m
c) Option a: a(2.0 s) = -0.25 m/s^2, x(2.0 s) = 2.75 m; a(5.0 s) = -0.25 m/s^2, x(5.0 s) = 7.125 m
d) Option a: a(2.0 s) = 0.25 m/s^2, x(2.0 s) = 3.25 m; a(5.0 s) = 0.25 m/s^2, x(5.0 s) = 6.875 m

Answer 1

The acceleration of the particle at t = 2.0 s and t = 5.0 s is -0.25 m/s². The position of the particle at t = 2.0 s is 3.25 m and at t = 5.0 s is 5.875 m.

Step-by-step explanation:

The acceleration of the particle at t = 2.0 s and t = 5.0 s can be found by taking the derivative of the velocity function. The derivative of v(t) = A + Bt¯¹ is a(t) = -B(t¯²) = -0.25(t¯²). Substituting t = 2.0 s and t = 5.0 s into this equation gives a(2.0 s) = -0.25(2.0¯²) = -0.25 m/s² and a(5.0 s) = -0.25(5.0¯²) = -0.25 m/s².

The position of the particle at t = 2.0 s and t = 5.0 s can be found by integrating the velocity function. The integral of v(t) = A + Bt¯¹ is x(t) = At + B(ln|t|) + C, where C is a constant of integration. Given that x(t = 1 s) = 0, we can find C by substituting t = 1 s and x = 0 into the equation. This gives 0 = A(1) + B(ln|1|) + C = 2 + C, so C = -2. Therefore, the position of the particle at t = 2.0 s is x(2.0 s) = (2)(2.0) + (0.25)(ln|2.0|) - 2 = 4.0 + 0.25(0.693) - 2 = 3.25 m, and the position at t = 5.0 s is x(5.0 s) = (2)(5.0) + (0.25)(ln|5.0|) - 2 = 10.0 + 0.25(1.609) - 2 = 5.875 m.