A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is:
(2006)
Concept: Average velocity is total displacement over total time. For a complete circular lap, displacement is zero, so average velocity is \( 0 \). Average speed is total distance over total time. Total distance is circumference \( 2\pi R = 2 \times 3.14 \times 100 = 628 \text{ m} \). Total time is \( 62.8 \text{ s} \). Average speed \( = 628/62.8 = 10 \text{ m/s} \).
A particle moves along a straight line OX. At a time \( t \) (in seconds) the distance \( x \) (in meters) of the particle from O is given by \( x = 40 + 12t – t^3 \). How long would the particle travel before coming to rest?
(2006)
Concept: Particle comes to rest when velocity is zero. Velocity \( v = \frac{dx}{dt} = 12 - 3t^2 \). Setting \( v=0 \) gives \( 12 - 3t^2 = 0 \), so \( t=2 \text{ s} \). Initial position at \( t=0 \) is \( x(0) = 40 \text{ m} \). Position at \( t=2 \) s is \( x(2) = 40 + 12(2) - (2)^3 = 56 \text{ m} \). Distance traveled is \( |x(2) - x(0)| = |56 - 40| = 16 \text{ m} \).
The displacement \( x \) of a particle varies with time \( t \) as \( x = ae^{-\alpha t} + be^{\beta t} \), where \( a, b, alpha \) and \( beta \) are positive constants. The velocity of the particle will
(2005)
Concept: Velocity is the time derivative of displacement. Calculate \( v = \frac{dx}{dt} = -a\alpha e^{-\alpha t} + b\beta e^{\beta t} \). The term \( -a\alpha e^{-\alpha t} \) decreases in magnitude (approaching zero), while the term \( b\beta e^{\beta t} \) increases exponentially. Thus, the velocity of the particle will go on increasing with time.
For a particle displacement time relation is \( t = \sqrt{x} + 3 \). Its displacement when its velocity is zero:
(1999)
Concept: Velocity is the time derivative of displacement. First, express \( x \) as a function of \( t \): from \( t = \sqrt{x} + 3 \), we get \( \sqrt{x} = t - 3 \), so \( x = (t-3)^2 \). Then find velocity \( v = \frac{dx}{dt} = 2t-6 \). Set \( v=0 \) to find when it is at rest: \( 2t-6=0 \) implies \( t=3 \text{ s} \). Substitute \( t=3 \text{ s} \) back into the displacement equation: \( x(3) = (3-3)^2 = 0 \text{ m} \).
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)
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:
Motion of a particle is given by equation \(S = 3t^3 + 7t^2 + 14t + 8\text{m}\). The value of acceleration of the particle at \(t = 1 \text{ sec}\) is:
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)
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.