Physics, asked by Anonymous, 7 months ago

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Answered by EuphoricEpitome
11

Given :

Case 1:

train starts from rest (u = 0)

accelerates uniformly at = 2 m/s² ( a = 2)

for 10 s (t = 10s)

Case 2:

Train maintains constant speed for 200 s

Case 3:

breaks are applied , and uniformly retarded

comes to rest in 50 s.

To find :

i) Maximum velocity reached

ii) Retardation in last 50 s

iii)Total distance travelled

iv) Average velocity of train

Solution :

I) In case 1:

u = 0 , a = 2 m /s² , t = 10 s

to find :

v (final velocity)

we know that,

\bold{v = u + at}

by putting the values

v = 0 + 2 × 10

v = 20 m/s

in case 2:

the train maintains constant speed so velocity = 20 m/s

in case 3:

v = 0 (as it comes to rest)

therefore maximum velocity = 20 m/s

ii) u = 20 m/s , v = 0 , t = 50 s

v = u + at

0 = 20 + 50 a

20 = -50 a

 a = \frac {\cancel{-20}}{\cancel{50}} \\ \\ \\  a = \frac{-2}{5} \\ \\ \\ a = -0.4 \: m/s^2

* Note - negative sign means it is retardation .

iii) We know that,

\bold{s = ut + \frac{1}{2} at^2} \\ \\ \\In \: case \: 1 \\ \\ \\ u = 0 , a = 2, t = 10 \\ \\ by\: putting\:values \\ \\ \\ s = 0 + \frac{1}{\cancel{2}}\times \cancel{2} \times 10 \times 10 \\ \\ \\ \bold{\therefore \:s = 100 m}

in case 2:

constant speed for 200 s

We know that - acceleration = 0 , when velocity is constant .

therefore , u = 20 m/s , a = 0 , t = 200 s

\bold{s = ut + \frac{1}{2} at^2} \\ \\ \\ by\: putting\: values \\ \\ \\ s = (20 \times 200) + \frac{1}{2} \times 0 \times 20 \times 20 \\ \\ \\ s = 4000 \: m

in case 3:

u = 20 m/s , v = 0 , a = -0.4

We know that,

= + 2as

by putting the values

0² = 20² + 2(-0.4) s

0 = 400 -0.8 s

400 = 0.8 s

 s =  \frac{400}{0.8} \\ \\ \\ = \frac{4000}{8} \\ \\ \\ \therefore \: s = 500 \: m

therefore ,

total distance = 100 + 4000 + 500 = 4600 m .

iv). Average speed =  \frac{Total\: distance}{Total\:time\:taken} \\ \\ \\ = \frac{4600}{10 + 200 + 50} \\ \\ \\ = \frac{\cancel{4600}}{\cancel{260}}  = 17.69 \\ \\ \\ \bold{Average\:speed = 17.69 m/s }

Answered by aishi2020
12

Answer:

I did (i) & (iii)

Mark the above answerer as brainliest as she did all

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