A small particle moving with a uniform acceleration a, covers distances X and Y in the first two equal and consecutive intervals of time t. Show that a = (Y-X) / t²

I tried simplifying this using the 3 equations of motion, but the answer I'm getting is quite different from what was required to prove
Cc MrShape Femi4 Martinez39s

Let the two equal time intervals for which the particle covered distances X and Y be [ t₁ , t₂ ] and [ t₁ , t₂ ] respectively. It follows that
T = t₂ - t₁ = t₃ - t₂ ——— (1)

The antiderivative of a(t) = a with respect to t is v(t) = at + C. C is the initial velocity. Since X is the distance traveled in the first interval, X is the definite integral of v(t) with respect to t from t₁ to t₂. Similarly, Y is the definite integral of v(t) with respect to t from t₂ to t₃. When you solve the integrals rightly using t₁ , t₂ , & t₃ as the limits of integration, you have

X = ½a(t₂)² + Ct₂ - ½a(t₁)² - Ct₁.
Y = ½a(t₃)² + Ct₃ - ½a(t₂)² - Ct₂.

We you solve Y - X and simplify using eq (1), you get

Y - X = ¼a(t₃ - t₂)² = ¼a(2T)²

_{NB: In your question, T is written as t.}

Solved

Nice one boss

Martinez39s:
Let the two equal time intervals for which the particle covered distances X and Y be [ t₁ , t₂ ] and [ t₁ , t₂ ] respectively. It follows that
T = t₂ - t₁ = t₃ - t₂ ——— (1)

The antiderivative of a(t) = a with respect to t is v(t) = at + C. C is the initial velocity. Since X is the distance traveled in the first interval, X is the definite integral of v(t) with respect to t from t₁ to t₂. Similarly, Y is the definite integral of v(t) with respect to t from t₂ to t₃. When you solve the integrals rightly using t₁ , t₂ , & t₃ as the limits of integration, you have

X = ½a(t₂)² + Ct₂ - ½a(t₁)² - Ct₁.
Y = ½a(t₃)² + Ct₃ - ½a(t₂)² - Ct₂.

We you solve Y - X and simplify using eq (1), you get

Y - X = ¼a(t_³ - t₂)² = ¼a(2T)²

_{NB: In your question, T is written as t.}

Nice
Further conclude it

Solution:

distance (d1) = X
distance (d2) = Y
time = t1 = t2 = t (equal and consecutive intervals of time)

initial velocity = d1/t = X/t
final velocity = d2/t = Y/t

Acceleration (a) = (final velocity - initial velocity)/t

Putting X/t and Y/t into the formula for acceleration

We have...

Acceleration = (Y/t - X/t)/t

Find the LCM inside the bracket

We... (Y-X)/t ÷ t
(Y-X)/t × 1/t

Acceleration = (Y - X) / t²

Proved and shown

Loris576:

Nice
Further conclude it

lol. Y - X = ¼a(4T²) = aT²
This means a² = (Y - X)/T²

Martinez39s:
lol. Y - X = ¼a(4T²) = aT²
This means a² = (Y - X)/T²

a**

Thanks very much everyone I was looking at the question from the wrong perspective

peppo4live:
Solution:

distance (d1) = X
distance (d2) = Y
time = t1 = t2 = t (equal and consecutive intervals of time)

initial velocity = d1/t = X/t
final velocity = d2/t = Y/t

Acceleration (a) = (final velocity - initial velocity)/t

Putting X/t and Y/t into the formula for acceleration

We have...

Acceleration = (Y/t - X/t)/t

Find the LCM inside the bracket

We... (Y-X)/t ÷ t
(Y-X)/t × 1/t

Acceleration = (Y - X) / t²

Proved and shown

I like this simple approach

femi4:
I like this simple approach

Thanks