Motion in a Straight Line

2 s

Step 1: At maximum height, the final velocity v = 0 m s⁻¹. Step 2: Given: u = +19.6 m s⁻¹ (upward), a = −g = −9.8 m s⁻² (deceleration due to gravity acting downward). Step 3: Use the first equation of motion: v = u + at → 0 = 19.6 + (−9.8) × t. Step 4: 9.8t = 19.6 → t = 19.6/9.8 = 2 s. The stone decelerates at 9.8 m s⁻² and reaches zero velocity after 2 seconds. Option 4 s is incorrect as that would be the total time for the round trip (going up and coming back down).

Displacement of the object

Step 1: In a v-t graph, the y-axis is velocity (m s⁻¹) and the x-axis is time (s). Step 2: The area of a rectangle under a horizontal portion of the graph = velocity × time = v × t = displacement (using s = vt for uniform motion). Step 3: Even for non-uniform motion, integrating v with respect to t (which is the area under the curve) gives the displacement: s = ∫v dt. Step 4: Therefore, the area under the v-t graph always gives displacement, not distance (unless the motion is always in one direction). Acceleration is given by the slope of the v-t graph, not the area.

100 m

Step 1: Convert initial velocity: u = 72 km h⁻¹ = 72 × (1000/3600) = 20 m s⁻¹. Step 2: Final velocity v = 0 m s⁻¹, time t = 10 s. Step 3: Use s = ((u + v)/2) × t = ((20 + 0)/2) × 10 = 10 × 10 = 100 m. Alternatively, find deceleration: a = (v − u)/t = (0 − 20)/10 = −2 m s⁻², then s = ut + ½at² = 20×10 + ½×(−2)×100 = 200 − 100 = 100 m. The option 720 m is wrong as it uses 72 km h⁻¹ directly without unit conversion. The option 200 m forgets to apply deceleration.

5 s

Step 1: Given: u = 15 m s⁻¹ (eastward), a = −3 m s⁻² (deceleration means negative acceleration), v = 0 m s⁻¹. Step 2: Use the first equation of motion: v = u + at. Step 3: 0 = 15 + (−3) × t → 3t = 15 → t = 5 s. Step 4: The car will stop after 5 seconds. The option 3 s is wrong (it comes from incorrectly computing 15/5 = 3). The option 45 s is incorrect as it multiplies 15 × 3 instead of dividing. Always check: after 5 s, velocity = 15 − 3×5 = 0 ✓