Question

a body moving with rest of uniform acceleration travels a distance 'X' in the first t seconds and travels the distance 'Y' with same acceleration in next 2t seconds,then (A)4x (B) 3x (C) 1/2x (D) 2x ?

Solve step by step please!​

Answer 1

Step-by-step explanation:

Answer :-

Correct relationship is Y = 3X [Option.B]

Explanation :-

We have :-

→ Initial velocity (u) = 0

→ Distance in first 't' seconds = X

→ Distance in next '2t' seconds = Y

→ Acceleration is same.

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1st case :-

By the 2nd equation of motion :-

s = ut + ½at²

⇒ X = 0(t) + ½ × a × t²

⇒ X = ½at²

⇒ X = 0.5at² ----(1)

2nd case :-

Distance travelled by the body in next '2t' seconds will be :-

⇒ Y = [0(t) + ½ × a × (2t)²] - 0.5t²

⇒ Y = [½ × a × 4t²] - 0.5at²

⇒ Y = 2at² - 0.5at²

⇒ Y = 1.5at² ---(2)

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On dividing eq.2 by eq.1, we get :-

⇒ Y/X = 1.5at²/0.5at²

⇒ Y/X = 3

⇒ Y = 3X

Answer 2

Correct relationship is Y = 3X [Option.B]

Explanation :-

We have :-

→ Initial velocity (u) = 0

→ Distance in first 't' seconds = X

→ Distance in next '2t' seconds = Y

→ Acceleration is same.

________________________________

1st case :-

By the 2nd equation of motion :-

s = ut + ½at²

⇒ X = 0(t) + ½ × a × t²

⇒ X = ½at²

⇒ X = 0.5at² ----(1)

2nd case :-

Distance travelled by the body in next '2t' seconds will be :-

⇒ Y = [0(t) + ½ × a × (2t)²] - 0.5t²

⇒ Y = [½ × a × 4t²] - 0.5at²

⇒ Y = 2at² - 0.5at²

⇒ Y = 1.5at² ---(2)

________________________________

On dividing eq.2 by eq.1, we get :-

⇒ Y/X = 1.5at²/0.5at²

⇒ Y/X = 3

⇒ Y = 3X