Physics, asked by serinayasmin79, 8 months ago

The motion of a particle is described by the equation x= 20cm+(4cm/ s^2)t^2 A) Find the displacement of the particle in the interval t=2 s and t’=5 s. B) Find the average velocity in this time interval C) Find instantaneous velocity at time t=2

Answers

Answered by AditiHegde
4

Given:

The motion of a particle is described by the equation x= 20cm+(4cm/ s^2)t^2

To find:

A) Find the displacement of the particle in the interval t=2 s and t’=5 s. B) Find the average velocity in this time interval C) Find instantaneous velocity at time t=2

Solution:

From given, we have,

An equation that represents the motion of a particle,

x = 20cm+(4cm/ s^2)t^2  ⇒ x = a + bt²

The general form of equation of motion is given by,

x = x₀ + V₀t + At²/2

where, A = acceleration and V₀ = initial velocity

Comparing the given equation with the standard equation, we have,

x₀ = 20, V₀ = 0 and A = 2b = 2(4) = 8

A) The displacement of the particle in the interval t=2 s and t’=5 s.

ΔS = x_{t2} - x_{t1} = a + b_{t2}² - a - b_{t1}² = b[{t2}² - {t1}²]

ΔS = 4 [5² - 2²] = 4 [25 - 4] = 4 [21] = 84

∴ The displacement of the particle in the interval t=2 s and t’=5 s is 84 cm.

B) The average velocity in this time interval

V_{a} = ΔS/Δt = ΔS / [t2 - t1] = 84/[5 - 2] = 84/3

∴ V_{a} = 28

∴ The average velocity in this time interval is 28 cm/s

C) Find instantaneous velocity at time t=2

The  equation of motion of a particle

V = V₀ + At

where, V₀ = 0 and A =  2b = 2 × 4 = 8 cm/s²

The instantaneous velocity at t = 2

Vi = At = 8 × 2 = 16

Therefore, the instantaneous velocity at time t=2 is 16 cm/s

Answered by CandyCakes
3

Answer:

Given:

The motion of a particle is described by the equation x= 20cm+(4cm/ s^2)t^2

To find:

A) Find the displacement of the particle in the interval t=2 s and t’=5 s. B) Find the average velocity in this time interval C) Find instantaneous velocity at time t=2

Solution:

From given, we have,

An equation that represents the motion of a particle,

x = 20cm+(4cm/ s^2)t^2  ⇒ x = a + bt²

The general form of equation of motion is given by,

x = x₀ + V₀t + At²/2

where, A = acceleration and V₀ = initial velocity

Comparing the given equation with the standard equation, we have,

x₀ = 20, V₀ = 0 and A = 2b = 2(4) = 8

A) The displacement of the particle in the interval t=2 s and t’=5 s.

ΔS = x_{t2} - x_{t1} = a + b_{t2}² - a - b_{t1}² = b[{t2}² - {t1}²]

ΔS = 4 [5² - 2²] = 4 [25 - 4] = 4 [21] = 84

∴ The displacement of the particle in the interval t=2 s and t’=5 s is 84 cm.

B) The average velocity in this time interval

V_{a} = ΔS/Δt = ΔS / [t2 - t1] = 84/[5 - 2] = 84/3

∴ V_{a} = 28

∴ The average velocity in this time interval is 28 cm/s

C) Find instantaneous velocity at time t=2

The  equation of motion of a particle

V = V₀ + At

where, V₀ = 0 and A =  2b = 2 × 4 = 8 cm/s²

The instantaneous velocity at t = 2

Vi = At = 8 × 2 = 16

Therefore, the instantaneous velocity at time t=2 is 16 cm/s

Explanation:

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