A train accelerates from rest at 2 m/s² for 10s then brakes at 4 m/s² — total distance

Question

A train starts from rest and accelerates at $2\ \text{m/s}^2$ for 10 seconds, then applies brakes and decelerates at $4\ \text{m/s}^2$ until it stops. Find the total distance travelled.

Solution — Step by Step

Phase 1 — Acceleration phase (0 to 10 s)

Given:

Initial velocity: $u_1 = 0$ (starts from rest)
Acceleration: $a_1 = +2\ \text{m/s}^2$
Time: $t_1 = 10\ \text{s}$

Find velocity at end of Phase 1 (this is the initial velocity for Phase 2): $v_1 = u_1 + a_1 t_1 = 0 + 2 \times 10 = 20\ \text{m/s}$

Distance during Phase 1: $s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2} \times 2 \times 100 = 100\ \text{m}$

Phase 2 — Braking phase (until train stops)

Given:

Initial velocity: $u_2 = 20\ \text{m/s}$ (from Phase 1)
Deceleration: $a_2 = -4\ \text{m/s}^2$ (negative because it's opposing motion)
Final velocity: $v_2 = 0$ (train stops)

Find distance during Phase 2: Using $v^2 = u^2 + 2as$ : $0 = (20)^2 + 2 \times (-4) \times s_2$ $0 = 400 - 8s_2$ $s_2 = \frac{400}{8} = 50\ \text{m}$

Find time for Phase 2 (bonus — for complete analysis)

Using $v = u + at$ : $0 = 20 + (-4) \times t_2$ $t_2 = \frac{20}{4} = 5\ \text{s}$

The train decelerates for 5 seconds to stop.

Total distance

$s_{total} = s_1 + s_2 = 100 + 50 = 150\ \text{m}$

Total distance travelled = 150 m

Verify using v-t graph logic

A velocity-time (v-t) graph for this motion would show:

A straight line rising from (0, 0) to (10, 20) — acceleration phase (slope = +2)
A straight line falling from (10, 20) to (15, 0) — braking phase (slope = -4)

The total displacement equals the area under the v-t graph (a trapezoid or triangle+triangle):

Area of Phase 1 triangle: $\dfrac{1}{2} \times 10 \times 20 = 100\ \text{m}$

Area of Phase 2 triangle: $\dfrac{1}{2} \times 5 \times 20 = 50\ \text{m}$

Total = 150 m ✓

Why This Works

The key principle: we treat the two phases completely independently because the equations of motion apply to any interval of uniform acceleration. Phase 1 has uniform positive acceleration; Phase 2 has uniform negative acceleration (deceleration). The end state of Phase 1 (velocity = 20 m/s) becomes the initial state of Phase 2.

The v-t graph method provides a geometric check — displacement is always the area under the v-t graph, regardless of whether the motion is accelerated or decelerated.

Alternative Method

We can also use energy methods for Phase 2. But kinematic equations are faster here since deceleration is uniform and the train stops (final KE = 0).

Alternatively: using $v^2 = u^2 + 2as$ directly:

For Phase 1: $v^2 = 0 + 2(2)(s_1) \Rightarrow s_1 = \dfrac{v^2}{4}$ . Here $v = 20$ , so $s_1 = 100$ m.

For Phase 2: $0 = v^2 + 2(-4)s_2 \Rightarrow s_2 = \dfrac{v^2}{8} = \dfrac{400}{8} = 50$ m.

💡 Expert Tip

Notice that in Phase 2, with twice the deceleration compared to acceleration, the braking distance is half the acceleration distance. This ratio (deceleration/acceleration = 4/2 = 2) means braking covers half the distance in half the time. This kind of proportional reasoning is useful for MCQ shortcuts in JEE Main.

Common Mistake

⚠️ Common Mistake

A very common error is using the total time in a single equation with average acceleration. You cannot use $s = ut + \frac{1}{2}at^2$ over both phases simultaneously because the acceleration is different in each phase. Always split the problem into constant-acceleration segments and analyse each separately.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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