Final answer:
The expression for the velocity of a particle moving in the horizontal direction is given as v(t) = A + Bt¯¹. To find the acceleration, we can take the derivative of the velocity function with respect to time. The position of the particle at specific times can be found by integrating the velocity function.
Explanation:
To find the expression for the velocity of a particle moving in the horizontal direction, we can use the information given in the question. The velocity function is given as v(t) = A + Bt¯¹, where A = 2 m/s, B = 0.25 m, and 1.0 s ≤ t ≤ 8.0 s. To determine the acceleration, we can take the derivative of the velocity function with respect to time. We can then find the position of the particle at specific times using the relationship between position and velocity.
Step 1: Find the acceleration:
To find the acceleration, we can take the derivative of the velocity function. The derivative of v(t) = A + Bt¯¹ with respect to time is a(t) = 0 - B(-1)t¯². Simplifying this expression gives us a(t) = Bt¯².
Step 2: Find the position at t = 2.0 s:
To find the position of the particle at t = 2.0 s, we integrate the velocity function from t = 1.0 s to t = 2.0 s. The integral of v(t) = A + Bt¯¹ with respect to time is x(t) = At + (B/2)t². Plugging in the values A = 2 m/s and B = 0.25 m, we can calculate x(2.0) as follows: x(2.0) = 2(2.0) + (0.25/2)(2.0)² = 4.0 + 0.25(4.0) = 4.0 + 1.0 = 5.0 m.
Step 3: Find the position at t = 5.0 s:
To find the position of the particle at t = 5.0 s, we integrate the velocity function from t = 1.0 s to t = 5.0 s. Using the same integral from step 2, plugging in the values A = 2 m/s and B = 0.25 m, we can calculate x(5.0) as follows: x(5.0) = 2(5.0) + (0.25/2)(5.0)² = 10.0 + 0.25(25.0) = 10.0 + 6.25 = 16.25 m.