Question

If a body moving with uniform acceleration in a straight line describes successive equal distances in time intervals(T1, T2, T3) respectively. Then prove 1/T1+1/T2+1/T3= 3/T1 +T2+T3​

Answer 1

Hello friend,

There's small correction. We have to prove 1/T1 - 1/T2 + 1/T3 = 3 / (T1 + T2 + T3).

◆ Proof -

Let v0, v1, v2 & v3 be velocities at time 0, after T1, after T1+T2 & after T1+T2+T3 respectively.

Now, average velocities in respective timeframes can be given by -

During T1, s/T1 = (v0+v1)/2 ...(1)

During T2, s/T2 = (v1+v2)/2 ...(2)

During T1, s/T3 = (v2+v3)/2 ...(3)

Average velocity during this time interval -

(s + s + s) / (T1 + T2 + T3) = (v0 + v3) / 2

3s / (T1 + T2 + T3) = (v0 + v3) / 2 ...(4)

Now, eqn.(1) - eqn.(2) + eqn.(3)

s/T1 - s/T2 + s/T3 = (v0+v1)/2 - (v1+v2)/2 + (v2+v3)/2

s (1/T1 - 1/T2 + 1/T3) = (vo + v1 - v1 - v2 + v2 + v3) / 2

s (1/T1 - 1/T2 + 1/T3) = (v0 + v3) / 2

From eqn.(4),

s (1/T1 - 1/T2 + 1/T3) = 3s / (T1 + T2 + T3)

1/T1 - 1/T2 + 1/T3 = 3 / (T1 + T2 + T3)

Thus proved. Best luck.

Answer 2

Answer:

1/T₁ - 1/T₂ + 1/T₃ = 3/(T₁+T₂+T₃)

Explanation:

If a body moving with uniform acceleration in a straight line describes successive equal distances in time intervals(T1, T2, T3) respectively. Then prove 1/T1 -1/T2+1/T3= 3/T1 +T2+T3​

Let say Velocities

V₀ , V₁ , V₂ & V₃  at T₀ , T₁ , T₁+T₂ & T₁+T₂+T₃ respectively

as body is moving with uniform acceleration

s = (V₀ + V₁)T₁/2

s = (V₁ + V₂)T₂/2

s = (V₂ + V₃)T₃/2

3s = (V₀ + V₃)(T₁+T₂+T₃)/2

=> V₀ + V₃ = 6s/(T₁+T₂+T₃)

V₀ + V₃ =  (V₀ + V₁) - (V₁ + V₂) + (V₂ + V₃)

=> V₀ + V₃ = 2s/T₁ - 2s/T₂ + 2s/T₃

=> 2s/T₁ - 2s/T₂ + 2s/T₃ =  6s/(T₁+T₂+T₃)

=> 1/T₁ - 1/T₂ + 1/T₃ = 3/(T₁+T₂+T₃)