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.

Step 1: Formula for average speed $v_avg = \dfrac{total distance}{total time}$.

Total distance $s = 50$ km.

Total time $t = 30$ min $= \dfrac{30}{60}$ h $= 0.5$ h.

Substitute: $v_avg = \dfrac{50}{0.5}$ km/h.

Simplify: $v_avg = 100$ km/h.

100 km/h

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

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

Step 1: Object accelerates at $a = 2$ m/s$^2$ for $t_1 = 5$ s (assume from rest).

Velocity after 5 s: $v = at_1 = 2 \times 5 = 10$ 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

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