Question

A train starts from rest and moves with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

Answer 1

Initial velocity, u = 0
Acceleration, a = 2 m/s²
Let the final velocity be v before the brakes are applied.
Now,
t = 30 s
v = u + at
v = 0 + 2 × 30
⇒ v = 60 m/s

(a) ${\mathrm{s}}_1=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_1=\frac{1}{2}\times 2\times {\left(30\right)}^2=900\mathrm{m}$
When the brakes are applied:
u' = 60 m/s
v' = 0
t = 1 min = 60 s
Acceleration:
$$\begin{gathered} a\text{'}=\frac{\left(v-u\right)}{t}=\frac{\left(0-60\right)}{60}=-1\mathrm{m}/{\mathrm{s}}^2 \\ {s}_2=\frac{v^2-u^2}{2a\text{'}}=\frac{0^2-{60}^2}{2\left(-1\right)}=1800\mathrm{m} \end{gathered}$$
s = s₁ + s₂ = 1800 + 900 = 2700 m
⇒ s = 2.7 km

(b) Maximum speed attained by the train, v = 60 m/s

(c) Half the maximum speed$=\frac{60}{2}=30\mathrm{m}/\mathrm{s}$
When the train is accelerating with an acceleration of 2 m/s²:

Distance, $\mathrm{s}=\frac{v^2-u^2}{2a\text{'}}$$=\frac{{30}^2-0^2}{2\times 2}$
⇒ s = 225 m

When the train is decelerating with an acceleration of $-$1 m/s²:
Distance, $\mathrm{s}=\frac{v^2-u^2}{2a\text{'}}$$=\frac{{30}^2-{60}^2}{2\left(-1\right)}$
⇒ s = 1350 m
Position from the starting point = 900 + 1350 = 2250
= 2.25 km