Question

a train starts from rest and accelerates uniformly for 30sec to acquire a velocity of 108kmph it travels with this velocity for 20min.the driver now applies breaks the train retards uniformly to stop after 20sec. find the total distance covered by the train?

Answer 1

initial velocity = 0
final velocity = 108 kmph (108 x 5 / 18 = 30 mps)
time taken by train to change its velocity = 30 seconds

a = (30-0)30
= 1 m/s²
distance travelled in 30 seconds =
$s = ut + \frac{1}{2} a {t}^{2} \\ = 0 + \frac{1}{2} a {t}^{2} \\ = \frac{1}{2} \times 1 \times {30}^{2} \\ = \frac{1}{2} \times 900 \\ = 450 \: metres \\ = \frac{450}{1000} \\ = 0.45 \: km$
train travels with a speed of 108 kmph for 20 minutes
then, distance covered in 20 minutes =
$= 108 \times \frac{20}{60} \\ = 108 \times \frac{1}{3} \\ = 36 \: km$
the retards and come to rest in 20 seconds.
that means,
final velocity becomes zero
initial velocity = 108 kmph (30 mps)

so acceleration = -3 / 2 m/s²

using third equation of uniformly accelerated motion,
$s = ut + \frac{1}{2} a {t}^{2} \\ = 30 \times 20 + \frac{1}{2} \times \frac{ - 3}{2} \times {20}^{2} \\ = 600 + \frac{ - 3}{4} \times 400 \\ = 600 + ( - 300) \\ = 600 - 300 \\ = 300 \: metres \\ = \frac{300}{1000 \: km} \\ = 0.300 \: km$
Total distance covered =
0.450 + 36 + 0.300
36 + 0.750
36.75 km

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Answer 2

Answer:

$36750 \: m$

Explanation:

When the train starts from rest,

u= 0,

t= 30s,

v= 108 km/h = 108 × 5/18 = 30 m/s.

Now,

a= v-u/t = 1 m/s².

Distance ,S1 = ut + 1/2 at²

= 0 + 1/2 × 1 × 30 × 30

= 450 m.

At uniform velocity of 30 m/s for 20 min, i.e., 1200s,

Distance, S2 = 30 × 1200 = 36000 m.

Now,

On applying brakes,

t= 20s,

u= 30 m/s,

v= 0.

a= v-u/t = 0-30/20 = -1.5 m/s²

Distance S3 = v² - u²/2a

= 0 - (30)² /2 × 1.5

= 300 m

Therefore,

The total distance traversed = S1 + S2 + S3

= (450 + 36000 + 300) m

= 36750 m