Braking bus: stopping distance, stopping time, and the distance not travelled

SPH3U Grade 11 Physics · Kinematics

A city bus is doing 79.2 km/h on a wet arterial road when the driver brakes hard and holds a steady deceleration of 5.50 m/s² until the bus stops. Find the distance the bus covers while braking, the time the stop takes, and how far the bus would have gone in that same time if the brakes had never been applied.

Step 1 of 3 · solve for Displacement

The speedometer says km/h and the deceleration is quoted in m/s², so the two givens are not yet in the same system — 79.2 ÷ 3.6 = 22.0 m/s fixes that. Then pick the equation with no t in it: speed, deceleration and distance are the only quantities in the question, so v² = v₀² + 2ad gets the braking distance without inventing a time first.

Rearranged for d

d = \tfrac{v^2 - v_0^2}{2 a}

v

v0

a

Your values, in your units

d = \tfrac{\left( 0\ \text{km/h} \right)^2 - \left( 79.2\ \text{km/h} \right)^2}{2 \, \left( -5.5\ \text{m/s}^{2} \right)}

Converted to base units

d = \tfrac{\left( 0\ \text{m/s} \right)^2 - \left( 22\ \text{m/s} \right)^2}{2 \, \left( -5.5\ \text{m/s}^{2} \right)}

Answer

d = 44

in

Carried onward at full precision, not this rounded figure.

Open the Velocity-Displacement Relation (v² = v₀² + 2ad) solver →

Step 2 of 3 · solve for Time

Now that the distance is known, a second equation supplies the time. Because the deceleration is uniform, the average speed over the stop is simply (79.2 + 0)/2 = 39.6 km/h, which is 11.0 m/s.

Rearranged for t

t = \frac{2 d}{v_0 + v}

d44 mcarried from step 1

v0

v

Your values, in your units

t = \frac{2 \, \left( 44\ \text{m} \right)}{\left( 79.2\ \text{km/h} \right) + \left( 0\ \text{km/h} \right)}

Converted to base units

t = \frac{2 \, \left( 44\ \text{m} \right)}{\left( 22\ \text{m/s} \right) + \left( 0\ \text{m/s} \right)}

Answer

t = 4

in

Carried onward at full precision, not this rounded figure.

Open the Displacement from Average Velocity solver →

Step 3 of 3 · solve for Distance

Same clock, no brakes: 79.2 km/h held for the time from step 2. Set this beside step 1 — it comes out exactly double, and the difference is the distance the brakes bought.

Rearranged for d

d = v t

v

t4 scarried from step 2

Your values, in your units

d = \left( 79.2\ \text{km/h} \right) \, \left( 4\ \text{s} \right)

Converted to base units

d = \left( 22\ \text{m/s} \right) \, \left( 4\ \text{s} \right)

Answer

d = 88

in

Carried onward at full precision, not this rounded figure.

Open the Speed, Distance & Time solver →

Why this order

The point of this chain is the choice in step 1. Three of the five kinematic equations could eventually produce the braking distance, but only v² = v₀² + 2ad does it without first solving for a time nobody asked about. Reading the question for what is missing — here, t — and picking the equation that omits it is the entire skill; students who instead reach for d = v₀t + ½at² end up solving a quadratic to answer a one-line question. Step 2 then deliberately switches equations, which is why the deceleration does not appear in it at all.

Step 3 exists to kill a stubborn intuition. Over the 4.00 s stop the bus averages 11.0 m/s, exactly half of the 22.0 m/s that 79.2 km/h works out to, so it covers exactly half the 88.0 m it would have covered at full speed — 44.0 m. That factor of two is not a coincidence of these numbers; it is what "uniform" deceleration means, and it holds for every constant-rate stop. The other half of the story is the squared term: braking distance grows as v₀², so the same bus at 118.8 km/h (33.0 m/s) needs 99 m, not 66 m. That is the arithmetic behind every posted speed limit near a school, and it is why the Ontario driver's handbook talks about stopping distance rather than stopping time.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.