A light-rail tram leaves a station from rest and accelerates smoothly at 1.25 m/s² until it reaches its cruising speed of 72.0 km/h. It holds that speed for 42.0 s along a straight reserved lane, then the driver brakes at a steady 1.60 m/s² and stops exactly at the next platform.
Given
v_c = 72 km/h — Cruising speed
a₁ = 1.25 m/s² — Acceleration leaving the platform
t_c = 42 s — Time spent cruising
a₂ = 1.6 m/s² — Braking deceleration (magnitude)
Determine
(a)the time the tram takes to reach cruising speed, and the distance that takes
(b)the distance covered while cruising
(c)the braking distance into the next station
(d)the time the braking takes
Step 1 of 5(a) · solve for Time
The speed board says km/h and the acceleration says m/s², so 72.0 ÷ 3.6 = 20.0 m/s happens before any equation is chosen. Then pick v = v₀ + at because it is the only kinematic equation with no distance in it — nothing about position is known yet.
Rearranged for t
t=av−v0
Your values, in your units
t=(1.25m/s2)(72km/h)−(0km/h)
Converted to base units
t=(1.25m/s2)(20m/s)−(0m/s)
Answer
t=16s
Carried onward at full precision, not this rounded figure.
Under uniform acceleration the average speed is exactly (0 + 20.0)/2 = 10.0 m/s — half the final speed, not the final speed. Multiplying 20.0 m/s by the 16.0 s from step 1 would double the true distance.
Rearranged for d
d=2v0+vt
16 scarried from step 1
Your values, in your units
d=2(0km/h)+(72km/h)(16s)
Converted to base units
d=2(0m/s)+(20m/s)(16s)
Answer
d=160m
Carried onward at full precision, not this rounded figure.
The cruise is the easy phase and students overthink it: with a = 0 the ½at² term is dead, and plain d = vt is not a shortcut — it is the whole law of constant-velocity motion.
Rearranged for d
d=vt
Your values, in your units
d=(72km/h)(42s)
Converted to base units
d=(20m/s)(42s)
Answer
d=840m
Carried onward at full precision, not this rounded figure.
Nobody asked how long the stop takes, so use the time-free equation v² = v₀² + 2ad. The deceleration goes in as −1.60 m/s² — feed it in positive and the distance comes out negative, which is the equation telling you the tram sped up into the platform.
Rearranged for d
d=2av2−v02
Your values, in your units
d=2(−1.6m/s2)(0km/h)2−(72km/h)2
Converted to base units
d=2(−1.6m/s2)(0m/s)2−(20m/s)2
Answer
d=125m
Carried onward at full precision, not this rounded figure.
Now the clock. The trap is dividing 125 m by 20.0 m/s to get 6.25 s — during the stop the tram averages only 10.0 m/s, so the answer is exactly twice that guess.
Rearranged for t
t=v0+v2d
125 mcarried from step 4
Your values, in your units
t=(72km/h)+(0km/h)2(125m)
Converted to base units
t=(20m/s)+(0m/s)2(125m)
Answer
t=12.5s
Carried onward at full precision, not this rounded figure.
Therefore the tram needs 16.0 s and 160 m to reach speed, cruises 840 m, and brakes to a stop over 125 m in 12.5 s — a 1125 m run between platforms, 70.5 s from door to door.
Why this order
The skill being drilled here is not algebra, it is selection: three phases, and for each one the question of which kinematic equation leaves out the quantity you neither know nor want. Part (a) has no distance in it, so v = v₀ + at; part (c) has no time in it, so v² = v₀² + 2ad; and the cruise in the middle is constant velocity, where four of the five equations collapse into d = vt. Students who instead grab d = v₀t + ½at² for the braking phase end up solving a quadratic for a time nobody asked about, then substituting it back — two chances to slip, zero new information.
The recurring number in this problem is ½, and it is worth seeing why it keeps appearing. Starting from rest, the tram averages half its final speed, so the 160 m of part (a) is exactly half of what 16.0 s at full speed would cover; braking to rest, it averages half its initial speed, so the 12.5 s of part (d) is exactly twice the naive 125/20 guess. Both are the same fact — uniform acceleration spends equal time at speeds below and above the midpoint — and it is also why the accelerating and braking phases together (285 m in 28.5 s) average exactly 10.0 m/s. Transit schedulers live on this arithmetic: closer stations mean a larger share of the trip spent in the half-speed phases, which is why an express line with distant stops needs no faster a tram to be dramatically faster door to door.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
