Question

What is the equation describing the position of particle b?

Answer 1

The equation describing the position of particle b is x(t) = 2t + 0.25ln(t) - 2, derived by integrating the given velocity function v(t) = A + Bt¯¹ and using the initial condition x(t = 1 s) = 0.

Step-by-step explanation:

To find the equation describing the position of particle b, we must integrate the velocity function. Given v(t) = A + Bt¯¹, where A = 2 m/s and B = 0.25 m, and given the time interval 1.0 s ≤ t ≤ 8.0 s, we need to perform the integration over time to get the position as a function of time, x(t).

First, we find the indefinite integral of the velocity function, v(t), which will give us an expression for x(t) plus a constant of integration, C. The integral is:

x(t) = ∫ (A + Bt¯¹) dt = At + Bln(t) + C.

Since we know that x(t = 1 s) = 0, we can use this condition to solve for C. We substitute t = 1 into the integral result and set x(1) to 0, resulting in C = -A - Bln(1) = -2 since the natural logarithm of 1 is 0.

So, the position function x(t) for particle b is:

x(t) = 2t + 0.25ln(t) - 2 (for 1.0 s ≤ t ≤ 8.0 s)

Answer 2

Let us suppose that two particles A and B start moving. Particle A starts moving at time t = 0 and velocity v. At time t, particle B has twice acceleration, half velocity and same position as the particle A has at time t=0.

So, position of particle B depends on time

The equation of motion can be written as: