Question

A body moving with uniform acceleration describes 10 metre in the 3rd sec and 12 metre in the 5th sec.

Find the distance traveled in (1) first 3 sec (2) 9th second of its motion.​

Answer 1

Answer :-

Distance travelled by the body :-

→ In first 3 sec is 27 m .

→ 9th second of it's motion is 16 m .

Explanation :-

Firstly, let's calculate the acceleration and initial velocity of the body.

For 3rd second :-

⇒ 10 = u + a/2[2 × 3 - 1]

⇒ 10 = u + 5a/2

⇒ 10 = u + 2.5a -----(1)

For 5th second :-

⇒ 12 = u + a/2[2 × 5 - 1]

⇒ 12 = u + 9a/2

⇒ 12 = u + 4.5a -----(2)

Subtracting eq.2 from eq.1 :-

⇒ 10 - 12 = u + 2.5a - u - 4.5a

⇒ - 2 = - 2a

⇒ a = 1 m/s²

Putting value of 'a' in eq.1 :-

⇒ 10 = u + 2.5(1)

⇒ u = 10 - 2.5

⇒ u = 7.5 m/s

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By using the 2nd equation of motion :-

s = ut + ½at²

⇒ s = 7.5(3) + ½ × 1 × (3)²

⇒ s = 22.5 + 4.5

⇒ s = 27 m

Now, from the equation of distance covered in the 'nth' second :-

sₙₜₕ = u + a/2(2n - 1)

⇒ s₉ₜₕ = 7.5 + 1/2[2 × 9 - 1]

⇒ s₉ₜₕ = 7.5 + 0.5[17]

⇒ s₉ₜₕ = 7.5 + 8.5

⇒ s₉ₜₕ = 16 m

Answer 2

Given :-

A  body moving with uniform acceleration describes 10 metre in the 3rd sec and 12 metre in the 5th sec

To Find :-

the distance traveled in (1) first 3 sec (2) 9th second of its motion.​

Solution :-

We know that

$\sf S_n = u + \dfrac{1}{2}a(2n-1)$

S₃ = u + 1/2a[2(3) - 1]

S₃ = u + a/2[6 - 1]

S₃ = u + a/2 × 5

S₃ = u + 5a/2

S₉ = u + 1/2a[2(5) - 1]

S₉ = u + 1/2a[10 - 1]

S₉ = u + a/2[9]

S₉ = u + 9a/2

On subtracting them

10 - 12 = u + 5a/2 - u + 9a/2

-2 = u + 5a - u - 9a/2

-2 = -4a/2

2 = 4a/2

2 = 2a

2/2 = a

1 = a

By putting in 2

12 = u + 9(1)/2

12 = u + 9/2

12 = u + 4.5

12 - 4.5 = u

7.5 = u

Now

s = ut + 1/2at²

s = 7.5 × 3 + 1/2 × 1 × (3)²

s = 22.5 + 1/2 × 9

s = 22.5 + 4.5

s = 27 m

In 9 th second

s = u + 1/2a [2n-1]

s = 7.5 + 1/2 × 1 [2(9) - 1]

s = 7.5 + 1/2 × 17

s = 7.5 + 8.5

s = 16 m