Question

A particle moving with constant acceleration describes in last second of its motion 9/25 of whole distance .If it starts from rest ,how long is the particle in motion and through what distance does it move if it describes 6cm in the first second?

Answer 1

Answer:

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Answer 2

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 0

Distance covered in 1 cm = 6 cm

Here, we get

Distance = 9d/25

We know that,

⇒ s = ut + 1/2at²

⇒ 6 = 1/2 × a × (1)²

⇒ a = 12 cm/s²

Now, the distance covered

⇒ d = 1/2at²

⇒ d = 1/2 × 12 × t²

⇒ d = 6t²...... (i)

Now distance covered in first second,

⇒ d₁ = 1/2 × a × (t - 1)²

⇒ d₁ = 6(t - 1)² cm .... (ii)

Now, for the whole journey,

⇒ 9d/25 = d - d₁

⇒ d₁ = d - 9d/25

⇒ d₁ = 16d/25

So, putting eq (i) and (ii) value, we get

⇒ 6(t - 1)² = 16/25 × 6t²

⇒ t - 1 = (4/5)t

⇒ 5(t - 1) = 4t

⇒ 5t - 5 = 4t

⇒ t = 5 sec

Hence, the time taken is 5 seconds.

Now, putting (i) value, we get

⇒ d = 6t²

⇒ d = 6 × (5)²

⇒ d = 6 × 25

⇒ d = 150 cm

Hence, the distance covered is 150 cm.