Question

Hyy guys!
Pls help me
Can anyone tell me why third equation of motion is used here..​

Answer 1

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1500m

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given,

final velocity = 90km/hr= 90*(5/18)= 25m/s.

time taken by train to attain that velocity is 2min,

i.e,120secs.

we can find acceleration by using,

a=(v-u)/t,

=>a= (25-0)/120= 5/24m/s^2.

then substitute a in S = Ut+1/2(at²),

to find the distance,

S = 0+ 1/2[(5/24)*(120*120)],

=>S = 1500m.

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