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NCERT Solutions for Class 9th Science Chapter 4 Chapter opener — Think It Over

Book page 48 Updated on2026-09-08

Q1.
How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
Answer

Enough distance to cover the distance our own vehicle needs to stop — and that has two separate parts.

stopping distance = reaction distance + braking distance
reaction distance = u × tr  (we keep moving while the driver reacts)
braking distance = u² / (2|a|)  (from v² = u² + 2as with v = 0)

Take a truck ahead and our car at u = 54 km h⁻¹ = 15 m s⁻¹, a driver reaction time tr ≈ 1 s and braking that gives |a| = 4 m s⁻² on a dry road:

reaction distance = 15 m s⁻¹ × 1 s = 15 m
braking distance = (15 m s⁻¹)² / (2 × 4 m s⁻²) = 225 / 8 m = 28.1 m
total = 15 m + 28.1 m = about 43 m
Why it happens: the truck ahead is also braking, so what really matters is the extra distance we need compared with it. Our reaction time is pure loss — for that whole second we travel at full speed with no braking at all. That is why the safe gap is usually quoted as a time gap (keep 2–3 seconds behind the vehicle ahead) rather than a fixed number of metres: a time gap automatically scales with speed.
Q2.
Does this distance depend upon the speed with which we are moving?
Answer

Yes — and much more strongly than most people expect, because the braking part grows as the square of the speed.

braking distance s = u² / (2|a|)
u = 15 m s⁻¹ (54 km h⁻¹)  →  s = 225 / 8 m = 28.1 m
u = 30 m s⁻¹ (108 km h⁻¹)  →  s = 900 / 8 m = 112.5 m

Doubling the speed does not double the braking distance — it makes it four times as long. The reaction distance (u × tr) only doubles, so at high speed the braking term dominates completely.

Tip: the same formula explains why speed limits fall so sharply on wet roads, near schools and in fog. A wet road lowers |a|, and s is inversely proportional to |a| — halve the braking capacity and the stopping distance doubles at every speed.
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