A particle moving in a straight line covers a half distance with a speed of 3m/s. The other two half of distance is covered in two equal intervals of time with speed of 4.5m/s and 7m/s. Find the average speed of the particle during this motion.
Answers
7+3+4.5/3=14.5/3=4.83m/s
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: