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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: Given

Initial velocity(u)= 0 m/s

Let distance in first 10 sec= s₁

Let distance in first 10 sec= s₂

a is acceleration, u is initial velocity and v is final velocity,

Step 2: Formula used

s=ut+12 a(t²) (2nd equation of motion)

v=u +at (1st equation of motion)

Step 3: Finding s₁

For first 10 seconds

s=ut+12 a(t²)

u=0 m/sec (body starts from rest)

s₁=0+12 a(t²)

s₁=12 a(10²)

s₁=50a

Step 4: Finding initial velocity for next 10 sec

For 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=at

v=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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