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 \( S = 0 ⇒ 6t - 1.5t^2 = 0 ⇒ t = 4\text{ s} \).
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: \( x(3) = (3-3)^2 = 0 \text{ m} \).
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 \(\vec{v}(t) = \frac{d\vec{r}}{dt} = 4\hat{i} + 4\that{j}\). At \(t = 1 \text{s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2} \text{m/s}\). The angle with the x-axis is \(tan\theta = \frac{v_y}{v_x} = \frac{4}{4} = 1 ⇒
\theta = 45^\circ\).
A particle is moving in x-y plane such that its x and y coordinates changes with time according to relation, \(x = 3t^2\) & \(y = 5t\) (here x & y are in m & t is in s). Speed of the particle at \(t = 2\) s, will be
The velocity components are \(v_x = \frac{dx}{dt} = 6t\) and \(v_y = \frac{dy}{dt} = 5\). At \(t = 2\) s, \(v_x = 12\text{ m/s}\) and \(v_y = 5\text{ m/s}\). Speed is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 5^2} = 13\text{ m/s}\).