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Please, Who Can Assist With The Solution To This Physics / Maths Problem by sureteeboy(m): 8:48am On Aug 12, 2020
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
Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by Martinez39s(m): 9:53am On Aug 12, 2020
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.

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by femi4: 10:16am On Aug 12, 2020
Solved

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by Loris576(m): 10:16am On Aug 12, 2020
Nice one boss

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by Loris576(m): 10:17am On Aug 12, 2020
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.
Nice
Further conclude it

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by peppo4live: 10:28am On Aug 12, 2020
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

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by Martinez39s(m): 10:38am On Aug 12, 2020
Loris576:

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

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by Loris576(m): 10:40am On Aug 12, 2020
Martinez39s:
lol. Y - X = ¼a(4T²) = aT²
This means a² = (Y - X)/T²
a**

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Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by sureteeboy(m): 11:08am On Aug 12, 2020
Thanks very much everyone grin I was looking at the question from the wrong perspective
Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by femi4: 12:08pm On Aug 12, 2020
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

3 Likes

Re: Please, Who Can Assist With The Solution To This Physics / Maths Problem by peppo4live: 3:45pm On Aug 12, 2020
femi4:
I like this simple approach

Thanks

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