Shatabdi express starts from rest and attains velocity of 180 km heigh

1.	Shatabdi express starts from rest and attains velocity of 180 km height in 2 minutes if acceleration in uniform find acceleration and displacement of the train
Answer» ANSWER: given, U = 0 m/s v = 180 * (5/18) m/s = 50 m/s t = 2 * 60 sec = 120 sec to FIND, a = ? s = ? formulas v = u + at _ _ _(1) s = ut + (1/2)a(t^2) _ _ _(2) v^2 = u^2 + 2as _ _ _(3) solution putting the VALUE of u, v and t in eq(1) v = u + at → 50 = 0 + a120 → 50/120 = a → a = 5/12 m/(s^2) putting the value of a and t in eq(2) s = ut + (1/2)a(t^2) →s = (0 120) + ((1/2) * (5/12) * 120 * 120) →s = 0 + 51060 →s = 3000m →s = 3 km

Shatabdi express starts from rest and attains velocity of 180 km height in 2 minutes if acceleration in uniform find acceleration and displacement of the train

Answer»

ANSWER:

given,

U = 0 m/s
v = 180 * (5/18) m/s = 50 m/s
t = 2 * 60 sec = 120 sec

to FIND,

a = ?
s = ?

formulas

v = u + at _ _ _(1)
s = ut + (1/2)a(t^2) _ _ _(2)
v^2 = u^2 + 2as _ _ _(3)

solution

putting the VALUE of u, v and t in eq(1)

v = u + at

→ 50 = 0 + a*120

→ 50/120 = a

→ a = 5/12 m/(s^2)

putting the value of a and t in eq(2)

s = ut + (1/2)a(t^2)

→s = (0 * 120) + ((1/2) * (5/12) * 120 * 120)

→s = 0 + 5*10*60

→s = 3000m

→s = 3 km

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