Verified answer

★ 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,

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

* Note - negative sign means it is retardation .

iii) We know that,

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

in case 3:

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

We know that,

v² = u² + 2as

by putting the values

0² = 20² + 2(-0.4) s

0 = 400 -0.8 s

400 = 0.8 s

therefore ,

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

iv). Average speed =

Answer:

I did (i) & (iii)

Mark the above answerer as brainliest as she did all