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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Hey Artillery, good to see you again. Here are the solutions to your assignment questions: 1. Speed-time graph analysis a) Find the speed V, if the deceleration in the first 3 seconds was 2 \, m/s^2. Step 1: Deceleration is the negative of the gradient of the speed-time graph. Deceleration = -change in speedchange in time 2 = -(V - 24)/(3 - 0) Step 2: Solve for V. 2 × 3 = -(V - 24) 6 = -V + 24 V = 24 - 6 V = 18 m/s b) Distance covered in the first 13 seconds. Step 1: The distance is the area under the speed-time graph. This consists of a trapezium (from t=0 to t=3) and a rectangle (from t=3 to t=13). Area of trapezium = (1)/(2) (u + v) h = (1)/(2) (24 + 18) × 3 Area of trapezium = (1)/(2) (42) × 3 = 21 × 3 = 63 \, m Step 2: Calculate the area of the rectangle. Area of rectangle = width × height = (13 - 3) × 18 Area of rectangle = 10 × 18 = 180 \, m Step 3: Sum the areas to find the total distance. Total distance = 63 + 180 = 243 m c) Value of t. Step 1: The car decelerates at 3 \, m/s^2 from t=13 until it comes to rest (speed 0 \, m/s). The initial speed for this segment is V = 18 \, m/s. Acceleration = change in speedchange in time -3 = (0 - 18)/(t - 13) Step 2: Solve for t. -3 (t - 13) = -18 t - 13 = (-18)/(-3) t - 13 = 6 t = 13 + 6 t = 19 s d) Total distance for the whole journey. Step 1: The total distance is the sum of the distance covered in the first 13 seconds (from part b) and the distance covered in the last segment (from t=13 to t=19). Distance in first 13 s = 243 \, m Step 2: Calculate the distance in the last segment (area of a triangle). Area of triangle = (1)/(2) × base × height = (1)/(2) × (19 - 13) × 18 Area of triangle = (1)/(2) × 6 × 18 = 3 × 18 = 54 \, m Step 3: Sum the distances. Total distance = 243 + 54 = 297 m e) Average speed of the car. Step 1: Average speed is calculated as total distance divided by total time. Average speed = Total distanceTotal time Average speed = 297 \, m19 \, s Average speed = 15.63 m/s (to 2 d.p.) f) Speed of the car at 15 seconds. Step 1: At t=15 seconds, the car is in the final deceleration phase, which starts at t=13 with a speed of 18 \, m/s and a deceleration of 3 \, m/s^2. Using the formula v = u + at', where u=18 \, m/s, a=-3 \, m/s^2, and t' = 15 - 13 = 2 \, s. v = 18 + (-3) × 2 v = 18 - 6 v = 12 m/s 2. If y = 2x^4 + 5, find (dy)/(dx). Step 1: Differentiate each term with respect to x. (dy)/(dx) = (d)/(dx)(2x^4) + (d)/(dx)(5) Step 2: Apply the power rule for differentiation ((d)/(dx)(ax^n) = anx^n-1) and the rule for constants ((d)/(dx)(c) = 0). (dy)/(dx) = 2 × 4x^4-1 + 0 (dy)/(dx) = 8x^3 3. Evaluate _0^2 (3x^2 + 2x)\,dx. Step 1: Find the indefinite integral of the function. (3x^2 + 2x)\,dx = 3x^2+12+1 + 2x^1+11+1 + C (3x^2 + 2x)\,dx = 3(x^3)/(3) + 2(x^2)/(2) + C (3x^2 + 2x)\,dx = x^3 + x^2 + C Step 2: Evaluate the definite integral using the limits of integration. [x^3 + x^2]_0^2 = (2^3 + 2^2) - (0^3 + 0^2) = (8 + 4) - (0 + 0) = 12 - 0 = 12 Send me the next one 📸