The ratio of the distance traveled by a freely falling body in the \(1^{text{st}}\,\text{ }2^{text{nd}}\,\text{ }3^{text{rd}}\) and \(4^{text{th}}\) second:
(2022)
Concept: Galileo's law of odd numbers for free fall.
Formula: Distance in \(n^{text{th}}\) second is \(S_n = \frac{g}{2}(2n - 1)\).
Solution: \(S_1:S_2:S_3:S_4 = (2(1)-1):(2(2)-1):(2(3)-1):(2(4)-1) = 1:3:5:7\).
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10\) seconds is \(S_1\) and that covered in the first \(20\)seconds is \(S_2\), then:
(2009)
For constant acceleration from rest, distance \(S = \frac{1}{2}at^2\). So \(S \propto t^2\). For \(t=10 \text{ s}\), \(S_1 = \frac{1}{2}a(10)^2 = 50a\). For \(t=20 \text{ s}\), \(S_2 = \frac{1}{2}a(20)^2 = 200a\). Therefore, \(S_2 = 4S_1\).
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \text{ m s}^{-2}\), in the third second is
(2008)
The distance in the \(n^{\text{th}}\) second is given by \(S_n = u + \frac{a}{2}(2n - 1)\). Here (u=0), \(a = \frac{4}{3} \text{ m s}^{-2}\) and (n=3). So \(S_3 = 0 + \frac{4/3}{2}(2 times 3 - 1) = \frac{2}{3}(5) = \frac{10}{3} \text{ m}\).
A particle moves in a straight line with a constant acceleration. It changes its velocity from \(10 \text{ m s}^{-1}\) to \(20 \text{ m s}^{-1}\) while passing through a distance \(135 \text{ m}\) in (t) second. The value of (t) is
(2008)
Given \(u = 10 \text{ m/s}\), \(v = 20 \text{ m/s}\), (\S = 135 \text{ m}\). Using \(v^2 = u^2 + 2aS\), \((20)^2 = (10)^2 + 2a(135) \Rightarrow 400 = 100 + 270a \Rightarrow a = \frac{300}{270} = \frac{10}{9} \text{ m/s}^2\). Now use \(v = u + at\), \(20 = 10 + \frac{10}{9}t \Rightarrow 10 = \frac{10}{9}t \Rightarrow t = 9 \text{ s}\).
The velocity of train increases uniformly from \(20 \text{ km/h}\) to \(60 \text{ km/h}\) in 4 hours. The distance travelled by the train during this period, is:
(1994)
Given initial velocity \(u = 20 \text{ km/h}\), final velocity \(v = 60 \text{ km/h}\), and time \(t = 4 \text{ h}\). For uniform acceleration, the distance \(S = \frac{u+v}{2}t\). Plugging in the values, \(S = \frac{20 + 60}{2} \times 4 = \frac{80}{2} \times 4 = 40 \times 4 = 160 \text{ km}\).
A particle has initial velocity \(2\hat{i} + 3\hat{j}\) and acceleration \(0.3\hat{i} + 0.2\hat{j}\) . The magnitude of velocity after 10 sec will be:
(2012 Pre)
Given \(\vec{v}_0 = 2\hat{i} + 3\hat{j}\), \(\vec{a} = 0.3\hat{i} + 0.2\hat{j}\), and \(t = 10\text{ s}\). Using \(\vec{v} = \vec{v}_0 + \vec{a}t\), we get \(\vec{v} = (2\hat{i} + 3\hat{j}) + (0.3\hat{i} + 0.2\hat{j})(10) = (2\hat{i} + 3\hat{j}) + (3\hat{i} + 2\hat{j}) = 5\hat{i} + 5\hat{j}\). The magnitude of velocity is \(|\vec{v}| = \sqrt{5^2 + 5^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}\).
A particle starting from the origin \((0, 0)\) moves in a straight line in the \((x, y)\) plane. Its coordinates at a later time are \((\sqrt{3}, 3)\). The path of the particle makes with the x-axis an angle of:
(2007)
The particle moves from \((0,0)\) to \((\sqrt{3}, 3)\). The slope of this straight line path is \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{\sqrt{3} - 0} = \frac{3}{\sqrt{3}} = \sqrt{3}\). The angle \(theta\) with the x-axis is given by \(tan\theta = m\). So, \(tan\theta = \sqrt{3}\) which implies \(\theta = 60^\circ\).
A truck moving with velocity \(36\text{ km/hr}\) is stopped by applying brakes in \(2\text{ s}\). If same truck moves with speed \(144\text{ km/hr}\) and brakes are applied then the stopping time will be (Assume the same retardation in both cases)
From \(v = u - at\), for stopping \(v = 0\), which gives \(t = u/a\). Since retardation \(a\) is constant, stopping time \(t \propto u\). Since the speed increases by a factor of \(144/36 = 4\), the stopping time becomes \(4 \times 2 = 8\text{ s}\).
With a constant acceleration along x-axis, a car covers a distance of \(20\text{ m}\) during fourth second of its motion and \(25\text{ m}\) during fifth second of its motion. Distance covered by it during third second of motion was
Using \(S_n = u + \frac{a}{2}(2n - 1)\), we get \(u + 3.5a = 20\) and \(u + 4.5a = 25\). Solving gives \(a = 5\text{ m/s}^2\) and \(u = 2.5\text{ m/s}\). Thus, \(S_3 = 2.5 + 2.5(5) = 15\text{ m}\).