Question

A particle starts from rest with a uniform acceleration
and travels 57 cm in the 10th second. Find out this dis-
tance travelled in 12 s and the velocity at that instant.
[432 cm; 72 cm/s​

Answer 1

Question:-

A particle starts from rest with a uniform acceleration and travels 57 cm in the 10th second. Find out this distance travelled in 12 s and the velocity at that instant.

Answer:-

Given:

Initial velocity = u = 0m/s

Acceleration = constant = a

Displacement in 10th second = s₁₀ = 57 cm

We know that, displacement in nth second = sₙₜₕ = u + (1/2)(a)(2n - 1)

Here it is given displacement in 10th second, so the value of n will be 10

So,

sₙₜₕ = u + (1/2)(a)(2n - 1)

→ s₁₀ = 0 + (1/2)(a)[2 * 10 - 1]

→ 57 = 0 + (1/2)(a)(19)

→ 57 = 9.5 * a

→ a = 6 m/s²

Distance travelled in 12 seconds :-

s = ut + (1/2)at²

→ s = (0 * 12) + (1/2)(6)(12)²

→ s = 0 + (3)(144)

→ s = 432 m

Final velocity at 12 seconds :-

v = u + at

→ v = 0 + (6)(12)

→ v = 72 m/s

Ans. 432m, 72 m/s

Note:-

• s = ut + (1/2)at² gives total displacement till that instant.
• sₙₜₕ = u + (1/2)(a)(2n - 1) gives only that amount of displacement covered at that instant of time.

Answer 2

Explanation:

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