Question.
A train is travelling at a speed of 90 kmh–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 ms–2. Find how far the train will go before it is brought to rest.
A train is travelling at a speed of 90 kmh–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 ms–2. Find how far the train will go before it is brought to rest.
Solution:
Given, initial speed of train,
$\mathrm{u}=90 \mathrm{~km} \mathrm{~h}^{-1}=90 \times \frac{5}{18}=25 \mathrm{~ms}^{-1}$
Final speed, $\mathrm{v}=0 \mathrm{~ms}^{-1}$,
Acceleration, $\mathrm{a}=-0.5 \mathrm{~ms}^{-2}$,
Distance covered, $\mathrm{s}=$ ?
Using the relation $v^{2}-u^{2}=2 a s$, we have
$s=\frac{v^{2}-u^{2}}{2 a}=\frac{0-(25)^{2}}{2 \times(-0.5)}=625 \mathrm{~m}$
Given, initial speed of train,
$\mathrm{u}=90 \mathrm{~km} \mathrm{~h}^{-1}=90 \times \frac{5}{18}=25 \mathrm{~ms}^{-1}$
Final speed, $\mathrm{v}=0 \mathrm{~ms}^{-1}$,
Acceleration, $\mathrm{a}=-0.5 \mathrm{~ms}^{-2}$,
Distance covered, $\mathrm{s}=$ ?
Using the relation $v^{2}-u^{2}=2 a s$, we have
$s=\frac{v^{2}-u^{2}}{2 a}=\frac{0-(25)^{2}}{2 \times(-0.5)}=625 \mathrm{~m}$
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