A body covers 10m in the second second and 25m in fifth second of its motion. If the motion is uniformly accelerated, how far will it go in seventh second?. Created by SharmaShivam. Physics

We know that Where,S is the displacement in nth second, a is the acceleration and n is the time and u is the initial velocity. Displacement in second second:-Displacement in Fifth second :-Since,body is uniformly accelerated. Thus, Both of the accelerations must be equal. Therefore :-Hence, Initial velocity = 2.5 m/secAcceleration will be :-Acceleration = 5 m/sec²Displacement in 7th second will be :-

A body covers 10m in the second second and 25m in fifth second of its motion. If...

skh2

Verified answer

We know that

$S_{n^{th}}=u+\dfrac{a(2n-1)}{2}$

Where,

S is the displacement in nth second, a is the acceleration and n is the time and u is the initial velocity.

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Displacement in second second:-

$S_2=u+\dfrac{a(4-1)}{2}\\ \\ \\S_2=u+\dfrac{3}{2}a\\ \\ \\ \dfrac{3}{2}a=10-u\\ \\ \\a=\dfrac{2(10-u)}{3}$

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Displacement in Fifth second :-

$S_5=u+\dfrac{a(10-1)}{2}\\ \\ \\S_5=u+\dfrac{9a}{2}\\ \\ \\a=\dfrac{2}{9}(25-u)$

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Since,body is uniformly accelerated.

Thus, Both of the accelerations must be equal.

Therefore :-

$\frac{2}{3}(10 - u) = \frac{2}{9}(25 - u) \\ \\ \\10 - u = \frac{(25 - u)}{3} \\ \\ \\3(10 - u) = 25 - u \\ \\ \\30 - 3u = 25 - u \\ \\ \\3u - u = 30 - 25 \\ \\ \\2u = 5 \\ \\ \\u = \frac{5}{2} \\ \\ \\u = 2.5$

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Hence,

Initial velocity = 2.5 m/sec

Acceleration will be :-

$a = \frac{2}{3}(10 - u) \\ \\ \\a = \frac{2}{3}(10 - 2.5) \\ \\ \\a = \frac{2 \times 7.5}{3} \\ \\ \\a = 5$

Acceleration = 5 m/sec²

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Displacement in 7th second will be :-

$S_7=u+\dfrac{a(2n-1)}{2}\\ \\ \\S_7=2.5+\dfrac{5*13}{2}\\ \\ \\S_7=2.5(1+13)\\ \\ \\S_7=2.5*14\\ \\ \\S_7=35\:\:m$

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pratyush4211

Verified answer

It cover 10 m in 2nd second of its Motion

We know

$s = ut + \frac{1}{2} at {}^{2}$

T =Time=2 seconds

S=10 m

$10 = u \times 2 + \frac{1}{2} a \times {2}^{2} \\ \\ 10 = 2u + 2a \\ \\ 10 = 2(u + a) \\ \\ u + a = \frac{10}{2} \\ \\ u + a = 5 \\ \\ a = 5 - u$

It cover 25 m in 5th second

We know

$s = ut + \frac{1}{2} {at}^{2} \\ \\$

S=25 m

T= Time given=5 seconds

$25 = u \times 5 + \frac{1}{2} a \times {5}^{2} \\ \\ 25 = 5u + \frac{25a}{2} \\ \\ 25 = \frac{10u + 25a}{2} \\ \\ 25 \times 2 = 10u + 25a \\ \\ 50 = 5(2u + 5a) \\ \\ \frac{50}{5} = 2u + 5a \\ \\ 10 = 2u + 5a \\ \\ 10 - 2u = 5a \\ \\ a = \frac{10 - 2u}{5}$