This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — A particle's acceleration is given by a = (6t - 18) m/s^2, and its initial velocity is V_0 = 24 m/s (at t=0). a) The velocity (V) of the particle at any time (t). Step 1: Integrate the acceleration function to find the velocity function. Velocity V(t) is the integral of acceleration a(t) with respect to time t. V(t) = a(t)\,dt V(t) = (6t - 18)\,dt V(t) = (6t^2)/(2) - 18t + C V(t) = 3t^2 - 18t + C Step 2: Use the initial condition to find the constant of integration C. Given that the initial velocity is 24 m/s at t=0: V(0) = 3(0)^2 - 18(0) + C = 24 C = 24 Step 3: Write the complete velocity function. V(t) = 3t^2 - 18t + 24 m/s The velocity of the particle at any time t is V(t) = 3t^2 - 18t + 24 m/s. b) The values of t at maximum displacement of the particle. Step 1: Maximum displacement occurs when the velocity of the particle is zero, as the particle momentarily stops before changing direction. Set V(t) = 0: 3t^2 - 18t + 24 = 0 Step 2: Solve the quadratic equation for t. Divide the entire equation by 3: t^2 - 6t + 8 = 0 Factor the quadratic equation: (t-2)(t-4) = 0 This gives two possible values for t: t-2 = 0 t = 2 s t-4 = 0 t = 4 s The values of t at maximum displacement are t=2 s and t=4 s. c) The distance covered by the particle between t=1 second and t=5 seconds. Step 1: Determine the displacement function s(t) by integrating the velocity function. s(t) = V(t)\,dt s(t) = (3t^2 - 18t + 24)\,dt s(t) = t^3 - 9t^2 + 24t + D_0 For calculating distance, the constant D_0 is not needed as it cancels out in differences. Step 2: Identify the intervals where the velocity changes sign. From part (b), we know V(t)=0 at t=2 s and t=4 s. These are the points where the particle changes direction. We need to calculate the distance in the intervals [1, 2], [2, 4], and [4, 5]. Step 3: Calculate the displacement in each interval and take the absolute value to find the distance. • For the interval [1, 2]: s(2) - s(1) = (2^3 - 9(2)^2 + 24(2)) - (1^3 - 9(1)^2 + 24(1)) s(2) - s(1) = (8 - 36 + 48) - (1 - 9 + 24) s(2) - s(1) = 20 - 16 = 4 m Distance for [1, 2] is |4| = 4 m. • For the interval [2, 4]: s(4) - s(2) = (4^3 - 9(4)^2 + 24(4)) - (2^3 - 9(2)^2 + 24(2)) s(4) - s(2) = (64 - 144 + 96) - (8 - 36 + 48) s(4) - s(2) = 16 - 20 = -4 m Distance for [2, 4] is |-4| = 4 m. • For the interval [4, 5]: s(5) - s(4) = (5^3 - 9(5)^2 + 24(5)) - (4^3 - 9(4)^2 + 24(4)) s(5) - s(4) = (125 - 225 + 120) - (64 - 144 + 96) s(5) - s(4) = 20 - 16 = 4 m Distance for [4, 5] is |4| = 4 m. Step 4: Sum the distances from each interval to find the total distance covered. Total distance = 4 m + 4 m + 4 m = 12 m. The distance covered by the particle between t=1 second and t=5 seconds is 12 m. Last free one today — make it count tomorrow, or type /upgrade for unlimited.