A car moving with a speed of \(50\text{ kmh}^{-1}\), can be stopped by brakes after at least \(6\text{ m}\). If the same car is moving at a speed of \(100\text{ kmh}^{-1}\), the minimum stopping distance is
Stopping distance \(s\) is proportional to \(u^2\). When the speed is doubled (from 50 to 100), the stopping distance increases by a factor of 4. Thus, \(s' = 4 \times 6 = 24\text{ m}\).
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 \(v = 3(4) - 24 + 3 = -9 \text{ ms}^{-1}\).
A train started from rest from a station and accelerated at \(2\text{ ms}^{-2}\) for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at \(4\text{ ms}^{-2}\) until it stopped at the next station. The distance between two station is
Max velocity \( v = 20 \text{ ms}^{-1}\). Acceleration distance \(s_1 = \frac{1}{2}(2)(10^2) = 100\text{ m}\). Constant velocity distance \(s_2 = 20 \times 30 = 600\text{ m}\). Deceleration distance \(s_3 = \frac{20^2}{2 \times 4} = 50\text{ m}\).
Total distance = 750 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 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 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 particle moves in x-y plane according to rule \(x = a \sin \omega t\) and \(y = a \cos \omega t\). The particle follows :
Squaring and adding the coordinates: \(x^2 + y^2 = a^2 \sin^2\omega t + a^2 \cos^2\omega t = a^2\). This is the standard equation of a circle of radius \(a\).
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.
A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is :
For uniform acceleration, the velocity midway between two points is given by the formula: \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{2500}{2}} = 25\sqrt{2}\text{ km/hr}\).
A particle moving with a velocity equal to 0.4 m/s is subjected to an uniform acceleration of
\( 0.15 m/s^{2} \) for 2 sec in a direction at right angles to its initial direction of motion. The resultant velocity is :
Initial velocity along x-axis is \(v_x = 0.4\text{ m/s}\). Velocity developed along the perpendicular y-axis is \(v_y = a_y t = 0.15 \times 2 = 0.3\text{ m/s}\). Resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\text{ m/s}\).
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 \(v = \frac{A}{B}(1 - e^{-Bt})\).