Question

11. Initial velocity of a particle moving along straight
line with constant acceleration is (20. + 2) m/s, if
its acceleration is a = (5 + 0.1) m/s2, then velocity
of particle with error, after time t = (10 + 1) s, is
(1) (70 + 3.1) m/s (2) (70 + 2) m/s
(3) (70 + 4) m/s (4) (70 + 8) m/s​

Answer 1

Explanation:

hope it helps u dear ...

Answer 2

Concept:

The discrepancy between a quantity's measured or inferred value and its actual value is known as the absolute error.

Given:

The initial velocity of the particle is (20 + 2)m/s and the acceleration of the particle is (5 + 0.1)m/s².

Find:

The velocity of the particle with error after time (10 + 1) second.

Solution:

The initial velocity of the particle, u = u + Δu = (20 + 2)m/s

The acceleration of the particle, a = a + Δa = (5 + 0.1)m/s²

The particle travels for a time, t = t +ΔtΔ = (10 + 1) s

Therefore, the true value of velocity is:

v = u + at

v = 20 + 5(10) = 70 m/s

The absolute error of the velocity of the particle is:

Δv = Δu + a(Δt) + t(Δa)

Δv = 2 + 5(1) + 10(0.1)

Δv = 2 + 5 + 1 = 8m/s

So, v = (70 +8)m/s

The velocity of the particle with error after time (10 + 1) s is (70 + 8)m/s.

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