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You're on a roll — a) To determine if the driver will be arrested, we need to check their speed. Police arrest drivers traveling above 80 km/hr. The speed for the first part of the journey was 60 km/hr. The speed for the second part of the journey was 80 km/hr. Since neither speed is above 80 km/hr (60 km/hr 80 km/hr and 80 km/hr 80 km/hr), the driver will not be arrested. The driver will not be arrested. b) We use the stopping distance formula d = (s^2)/(200) + (s)/(10). To find the stopping distance for s = 46 km/hr: Step 1: Substitute s = 46 into the formula. d = (46^2)/(200) + (46)/(10) Step 2: Calculate 46^2. d = (2116)/(200) + (46)/(10) Step 3: Perform the divisions. d = 10.58 + 4.6 Step 4: Add the values. d = 15.18 m The stopping distance for 46 km/hr is 15.18 m. To find the stopping distance for s = 85 km/hr: Step 1: Substitute s = 85 into the formula. d = (85^2)/(200) + (85)/(10) Step 2: Calculate 85^2. d = (7225)/(200) + (85)/(10) Step 3: Perform the divisions. d = 36.125 + 8.5 Step 4: Add the values. d = 44.625 m The stopping distance for 85 km/hr is 44.625 m. To find the speed when the stopping distance is 35 m: Step 1: Substitute d = 35 into the formula. 35 = (s^2)/(200) + (s)/(10) Step 2: Multiply both sides by 200 to eliminate the denominators. 35 × 200 = s^2 + (s)/(10) × 200 7000 = s^2 + 20s Step 3: Rearrange the equation into a standard quadratic form as^2 + bs + c = 0. s^2 + 20s - 7000 = 0 Step 4: Use the quadratic formula s = -b ± sqrt(b^2 - 4ac)2a, where a=1, b=20, c=-7000. s = -20 ± sqrt(20^2 - 4(1)(-7000))2(1) s = -20 ± sqrt(400 + 28000)2 s = -20 ± sqrt(28400)2 Step 5: Calculate the square root. sqrt(28400) ≈ 168.5229 Step 6: Calculate the two possible values for s. s_1 = (-20 + 168.5229)/(2) = (148.5229)/(2) ≈ 74.26145 s_2 = (-20 - 168.5229)/(2) = (-188.5229)/(2) ≈ -94.26145 Step 7: Choose the positive value for speed, as speed cannot be negative. s ≈ 74.26 km/hr The speed at which the car is traveling if its stopping distance is 35 m is 74.26 km/hr. Send me the next one 📸