The relation between times t and distance x is \(t = ax^2 + bx\), where a and b are constants. The acceleration is
Differentiating \(t = ax^2 + bx\) with respect to \(t\) gives \(1 = (2ax + b)v\). Differentiating again with respect to \(t\) gives \(a = \frac{dv}{dt} = -2av^3\).
If the velocity of a particle is given by \(v = (180 – 16x)^{1/2}\text{ ms}^{-1}\), then its acceleration will be
We have \(v^2 = 180 - 16x\). Comparing with \(v^2 = u^2 + 2ax\), we get \(2a = -16 ⇒ a = -8\text{ ms}^{-2}\).
A particle moves along a straight line such that its displacement at any time t is given by \(s = t^3 – 6t^2 + 3t + 4\). The velocity when its acceleration is zero is
Velocity \(v = 3t^2 - 12t + 3\) and acceleration \(a = 6t - 12\). Acceleration is zero at t = 2 s. Substituting \(t = 2\) in velocity equation gives \(v = 3(4) - 24 + 3 = -9 \text{ ms}^{-1}\).
The position of a particle is given by \(\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}\) where \(t\) is in seconds and \(r\) in meter. Find the magnitude and direction of velocity \(v(t)\), at \(t = 1\text{ s}\), with respect to x-axis.
Velocity is \(\vec{v} = \frac{d\vec{r}}{dt} = 4\hat{i} + 4t\hat{j}\). At \(t = 1\text{ s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Thus, magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2}\text{ ms}^{-1}\) and angle \(\tan\theta = \frac{4}{4} = 1 ⇒ \theta = 45^\circ\) with respect to the x-axis.
The x and y co-ordinates of a particle at any time \(t\) are given by : \(x = 7t + 4t^2\) and \(y = 5t\) where \(x\) and \(y\) are in m and \(t\) in s. The acceleration of the particle at 5 s is :-
Velocity components: \(v_x = \frac{dx}{dt} = 7 + 8t\) and \(v_y = \frac{dy}{dt} = 5\). Acceleration components: \(a_x = \frac{dv_x}{dt} = 8\) and \(a_y = \frac{dv_y}{dt} = 0\). Total acceleration is \(a = \sqrt{a_x^2 + a_y^2} = 8\text{ m/s}^2\) at any time.
The motion of a body falling from rest in a viscous medium is described by \(\frac{dv}{dt} = A – Bv\), where A and B are constants. The velocity at time t is given by :
Integrating the equation \(\int_0^v \frac{dv}{A - Bv} = \int_0^t dt\) gives \(-\frac{1}{B} \ln(\frac{A - Bv}{A}) = t\). Simplifying for velocity gives \(v = \frac{A}{B}(1 - e^{-Bt})\).
A particle moves along the parabolic path \(y = ax^2\) in such a way that x-component of velocity remains constant, say c. The acceleration of the particle is :
Given \(v_x = \frac{dx}{dt} = c\) (constant), hence \(a_x = 0\). Differentiating \(y = ax^2\) gives \(v_y = 2ax\frac{dx}{dt} = 2axc\). Differentiating again, \(a_y = 2ac\frac{dx}{dt} = 2ac^2\). Therefore, total acceleration vector is \(a = 2ac^2\hat{j}\).
The displacement \(s\) of a point moving in a straight line is given by : \(s = 8t^2 + 3t – 5\) where \(s\) is in cm and \(t\) in s. The initial velocity of the particle is :
Velocity is given by differentiating displacement with respect to time: \(v = \frac{ds}{dt} = 16t + 3\). For initial velocity, substitute \(t = 0\), which gives \(v = 3\text{ cm/s}\).
A point moves linearly such that \(\frac{dv}{dt} = -\alpha v^{1/2}\), where \(\alpha\) is a positive constant. At start, \(v = v_0\). Find time taken by particle to stop?
Integrating \(v^{-1/2} dv = -\alpha dt\) from \(v_0\) to \(0\) gives \(2\sqrt{v_0} = \alpha t\), so the time taken to stop is \(t = \frac{2\sqrt{v_0}}{\alpha}\).
A particle moves in a straight line with velocity \(v = (4t – 2)\text{ m/s}\). If the particle starts from \(x = 0\), then the position of particle at which it momentarily comes to rest is
The particle comes to rest when \(v = 0\), giving \(4t - 2 = 0 ⇒ t = 0.5\text{ s}\). Integrating velocity to find position, \(x = int_0^t (4t - 2)dt = 2t^2 - 2t\). At \(t = 0.5\text{ s}\), \(x = 2(0.5)^2 - 2(0.5) = -0.5\text{ m}\).