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Question

# A body starts from rest and moves with a constant acceleration. It travels a distance s₁ in first 10 s and a distance s₂ in next 10 s. Find the relation b/w s₁ and s₂.

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Solution

## Step 1: GivenInitial velocity(u)= 0 m/sLet distance in first 10 sec= s₁Let distance in first 10 sec= s₂a is acceleration, u is initial velocity and v is final velocity,Step 2: Formula useds=ut+12 a(t²) (2nd equation of motion)v=u +at (1st equation of motion)Step 3: Finding s₁For first 10 secondss=ut+12 a(t²) u=0 m/sec (body starts from rest) s₁=0+12 a(t²) s₁=12 a(10²) s₁=50aStep 4: Finding initial velocity for next 10 secFor next 10 seconds now initial velocity for s₂ is the final velocity of s₁ let final velocity of s₁=initial velocity of s₂=v so, v=u +at v=atv=10a (since u=0 and t=10)Step 5: Finding s₂ now s₂=ut+12 a(t²) s₂=(10a)10+12a(10²) s₂=100a+50a s₂=150a now Divide s₁ and s₂ s₁/s₂=50a150a =13 Hence, the relation b/w s₁ and s₂ is 3s₁=s₂

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