A particle of unit mass undergoes one dimensional motion such that its velocity varies according to \(v(x) = \beta x^{-2n}\) where \(\beta\) and (n) are constants and (x) is the position of the particle. The acceleration of the particle as a function of (x), is given by:
(2015)
1. \(-2n\beta^2 x^{-4n-1}\)
2. \(-2n\beta^2 x^{-2n+1}\)
3. \(-2n\beta^2 e^{-4n+1}\)
4. \(-2n\beta^2 x^{-2n-1}\)
View Answer
Given \(v = \beta x^{-2n}\). Acceleration \(a = v \frac{dv}{dx}\). First find \(\frac{dv}{dx} = \beta (-2n)x^{-2n-1}\). Then \(a = (\beta x^{-2n})(-2n\beta x^{-2n-1}\) = \(-2n\beta^2 x^{-4n-1}\).
The motion of a particle along a straight line is described by equation: \\(x = 8 + 12t – t^3\) where (x) is in metre and (t) in second. The retardation of the particle when its velocity becomes zero, is:
(2012 Pre)
1. \(24 m s^{-2}\)
2. (Zero)
3. \(6 m s^{-2}\)
4. \(12 m s^{-2}\)
View Answer
Given \(x = 8 + 12t - t^3\). Velocity \(v = \frac{dx}{dt} = 12 - 3t^2\). Acceleration \(a = \frac{dv}{dt} = -6t\). When (v=0), \(12 - 3t^2 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2 \text{ s}\). At \(t=2 \text{ s}\), \(a = -6(2) = -12 \text{ m/s}^2\). Retardation is \(-a = 12 \text{ m/s}^2\).
A particle moves a distance (x) in time (t) according to equation \(x = (t + 5)^{-1}\). The acceleration of particle is proportional to:
(2010 Pre)
1. \(\text{Velocity}^{2/3}\)
2. \(\text{Velocity}^{3/2}\)
3. \((\text{Distance})^2\)
4. \((\text{Distance})^{-2}\)
View Answer
Given \(x = (t + 5)^{-1}\). Velocity \(v = \frac{dx}{dt} = -(t + 5)^{-2}\). Acceleration \(a = \frac{dv}{dt} = 2(t + 5)^{-3}\). From \(v = -(t + 5)^{-2}\), we have \((t+5)^{-1} = ((-v)^{-1/2})\). So \(a = 2((t+5)^{-1})^3 = 2((-v)^{-1/2})^3 = 2(-v)^{3/2}\). Thus, \(a \propto (\text{Velocity})^{3/2}\).
The ‘x’ and ‘y’ coordinates of the particle at any time are \(x = 5t – 2t^2\) and \(y = 10t\), respectively, where ‘x’ and ‘y’ are in metres and ‘t’ in seconds. The acceleration of the particle at \(t = 2\text{ s}\) is:
(2017-Delhi)
1. \(5\text{ m/s}^2\)
2. \(-4\text{ m/s}^2\)
3. \(-8\text{ m/s}^2\)
4. \(0\)
View Answer
Given \(x = 5t - 2t^2\) and \(y = 10t\). Differentiating twice with respect to time to find acceleration. \(v_x = \frac{dx}{dt} = 5 - 4t\), \(a_x = \frac{dv_x}{dt} = -4\text{ m/s}^2\). \(v_y = \frac{dy}{dt} = 10\), \(a_y = \frac{dv_y}{dt} = 0\text{ m/s}^2\). The acceleration vector is \(\vec{a} = -4\hat{i}\). The x-component of acceleration is \(-4\text{ m/s}^2\), which is constant.