Question

a particle starts from rest and move a distance of S1 with a constant acceleration in time t, if this particle travel a distance of S2 in next time t, show that S2=3S1​

Answer 1

Answer:

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Answer 2

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Let a be the constant acceleration of the particle. Thens=ut+21at2 or s1=0+21×a×(10)2=50aand s2=[0+21a(20)2]−50a=150a

∴s2=3s1

Alternatively:Let a be constant acceleration and

s=ut+21at2, then s1=0+21×a×100=50a

Velocity after 10 sec. is v=0+10a

So, s2=10a×10+21a×100=150a⇒s2=3s1

 Let a be constant acceleration, using s=ut+21at2

so distance coreved in first 10 seconds s1=0+21×a×100=50a

Velocity after 10 sec. is v=0+10a

So, distance covered in next 10 seconds s2=10a×10+21a×100=150a⇒s2=3s1

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