A particle moves along a straight line with velocity given by v = (6 – 3t) where v is in m/s and t in seconds. Determine when the particle returns to its starting point.
Displacement is \(s = \int v \, dt = 6t - 1.5t^2\). To return to the starting point, \(s = 0 \implies 6t - 1.5t^2 = 0 \implies t = 4\text{ s}\).
A particle begins to move along straight line where the acceleration \( (a) \) of the particle varies with displacement \( (x) \) according to relation, \( a = 5x \), then velocity of the particle varies with displacement as
Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).
A particle moves along a straight line with velocity given by \( v = (6 – 3t) \) where \( v \) is in \( \text{m/s} \) and \( t \) in seconds. Determine when the particle returns to its starting point.
Displacement is \( S = \int v \, dt = \int_0^t (6 - 3t) \, dt = 6t - 1.5t^2 \). Returning to the starting point means \( S = 0 ⇒ 6t - 1.5t^2 = 0 ⇒ t = 4\text{ s} \).
If the velocity of a particle is \( v = At + Bt^2 \), where A and B are constants, then the distance travelled by it between 1 s and 2 s is:
(2016 – I)
Concept: Distance is the definite integral of velocity. Integrate \( v = At + Bt^2 \) from \( t=1 \) to \( t=2 \). \( \int_{1}^{2} (At + Bt^2) dt = \left[ A\frac{t^2}{2} + B\frac{t^3}{3} \right]_{1}^{2} \). Evaluating this gives \( \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right) = \frac{3A}{2} + \frac{7B}{3} \).
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: \( x(3) = (3-3)^2 = 0 \text{ m} \).
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:
(2000)
Given \(S = 3t^3 + 7t^2 + 14t + 8\). Velocity \(v = \frac{dS}{dt} = 9t^2 + 14t + 14\). Acceleration \(a = \frac{dv}{dt} = 18t + 14\). At \(t=1 \text{ s}\), \(a = 18(1) + 14 = 32 \text{ m/s}^2\).
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 \(a = (\beta x^{-2n})(-2n\beta x^{-2n-1}\) = \(-2n\beta^2 x^{-4n-1}\).