The distance travelled is approximately equal to the area under the velocity-time graph. Estimating the areas of the trapeziums from the graph gives a total distance of approximately 45 km.
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Initial speed u = 54 km h⁻¹, final speed v = 36 km h⁻¹, and time = 36 s = 0.01 h. Average speed = 45 km h⁻¹. Distance = 45 × 0.01 = 0.45 km = 450 m.
Given u = 28 m s⁻¹, v = 0 and s = 98 m. Using v² = u² + 2as, we get a = −4 m s⁻². Time taken, t = (v − u)/a = 7 s.
Total distance = 200 + 200 = 400 km and total time = 3 + 2 = 5 h. Average speed = 400/5 = 80 km/h. Since displacement is zero, average velocity = 0 km/h.
Yes, the safe distance depends on the speed of the vehicle. As speed increases, the stopping distance increases. Therefore, a vehicle moving faster must maintain a greater distance from the vehicle ahead.