An object undergoes constant acceleration after starting from rest and then travels 5 m in the first second. Determine how far it will go in the next second.

Exercise 1 Step 1: Formula for average speed v_avg = total distancetotal time. Total distance s = 50 km. Total time t = 30 min = (30)/(60) h = 0.5 h. Substitute: v_avg = (50)/(0.5) km/h. Simplify: v_avg = 100 km/h. 100 km/h Exercise 2 Step 1: The aircraft flies in a straight line due north, so magnitude of displacement equals distance traveled. Magnitude of displacement = 900 km. 900 km Exercise 3 Step 1: For two cars accelerating along the same road, if the situation implies the distance between them is constant (as context suggests), they must have the same acceleration. B implies that the two cars have the same acceleration ("rate"). B Exercise 4 Step 1: Object accelerates at a = 2 m/s^2 for t_1 = 5 s (assume from rest). Velocity after 5 s: v = at_1 = 2 × 5 = 10 m/s. Step 2: "Then travels 5" is ambiguous—does not specify if acceleration continues, stops, or time/distance for next phase. Cannot determine distance without more info. D Exercise 5 Step 1: Arrow projected vertically upward at initial speed u = 100 m/s, air resistance ignored (g downward). At return to ground (same level), by symmetry of motion or conservation of energy, speed v = u. v^2 = u^2 + 2as, where s=0 net displacement vertically, so v = u = 100 m/s. 100 m/s