Question

A train starts from rest and accelerates uniformly for 30 s to acquire a velocity of 108 km/h. It travels with this velocity for 20 min. The driver now applies brakes and the train retards uniformly to stop after 20 s. Find the total distance covered by the train.​

Answer 1

$\huge\mathfrak\purple{Bonjour!!}$

$\huge\bold\pink{Solution:-}$

♨ Answer:-

36750 m.

♨ Step-by-step 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. [The required solution..]

✔✔✔✔✔✔✔✔

Hope it helps...❣❣❣

⭐❤✨♥⭐❤✨♥⭐

Be Brainly...✌✌✌

♣ WALKER ♠

Answer 2

Answer:

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 =

$\begin{gathered}\\ \sf s = ut + \frac{1}{2} a {t}^{2} \\ \\ \sf= 0 + \frac{1}{2} a {t}^{2} \\ \\ \sf= \frac{1}{2} \times 1 \times {30}^{2} \\ \\ \sf= \frac{1}{2} \times 900 \\ \\ \sf= 450 \: metres \\ \\ \sf= \frac{450}{1000} \\\\ \sf = 0.45 \: km\end{gathered}$

train travels with a speed of 108 kmph for 20 minutes

then, distance covered in 20 minutes =

$\begin{gathered}\\ \sf= 108 \times \frac{20}{60} \\ \\ \sf= 108 \times \frac{1}{3} \\\\ \sf = 36 \: km\end{gathered}$

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,

$\begin{gathered}\\ \sf \: s = ut + \frac{1}{2} a {t}^{2} \\\\ \sf = 30 \times 20 + \frac{1}{2} \times \frac{ - 3}{2} \times {20}^{2} \\ \\ \sf= 600 + \frac{ - 3}{4} \times 400 \\ \\ \sf= 600 + ( - 300) \\\\ \sf = 600 - 300 \\ \\ \sf \: = 300 \: metres \\ \\ \sf= \frac{300}{1000 \: km} \\ \\ \sf \: = 0.300 \: km\end{gathered}$

Total distance covered =

0.450 + 36 + 0.300

36 + 0.750

36.75 km

PLZ Mark It As Brainliest Answer.