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)
1. \(9\sqrt{2}\text{ units}\)
2. \(5\sqrt{2}\text{ units}\)
3. \(5\text{ units}\)
4. \(9\text{ units}\)
View Answer
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 has initial velocity \(3\hat{i} + 4\hat{j}\) and has acceleration \(0.4\hat{i} + 0.3\hat{j}\). Its speed after 10 s is:
(2010 Pre)
1. \(10\text{ units}\)
2. \(7\text{ units}\)
3. \(7\sqrt{2}\text{ units}\)
4. \(8.5\text{ units}\)
View Answer
Initial velocity \(vec{v}_0 = 3\hat{i} + 4\hat{j}\). Acceleration \(\vec{a} = 0.4\hat{i} + 0.3\hat{j}\). Time \(t = 10\text{ s}\). Using \(\vec{v} = \vec{v}_0 + \vec{a}t\), we get \(\vec{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j})(10) = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) = 7\hat{i} + 7\hat{j}\). Speed is the magnitude of velocity: \(|\vec{v}| = \sqrt{7^2 + 7^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}\).
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)
1. \(45^\circ\)
2. \(60^\circ\)
3. \(0^\circ\)
4. \(30^\circ\)
View Answer
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 \(\theta = 60^\circ\).
Two boys are standing at the ends A and B of a ground, where \(AB = a\). The boy at B starts running in a direction perpendicular to AB with velocity \(v_1\). The boy at A starts running simultaneously with velocity \(v\) and catches the other boy in a time t, where t is:
(2005)
1. \(\frac{a}{\sqrt{v^2+v_1^2}}\)
2. \(\frac{a}{\sqrt{v^2-v_1^2}}\)
3. \(a/(v-v_1)\)
4. \(a/(v+v_1)\)
View Answer
Let B be at \((0,0)\) and A at \((a,0)\) at \(t=0\). Boy B's position at time \(t\) is \(\vec{r}_B = v_1 t \hat{j}\). Boy A moves with velocity \(\vec{v}_A = v_{Ax}\hat{i} + v_{Ay}\hat{j}\). For A to catch B, their positions must be equal at time \(t\). So, \(a\hat{i} + \vec{v}_A t = v_1 t \hat{j}\). This implies \(v_{Ax} = -a/t\) and \(v_{Ay} = v_1\). The magnitude of A's velocity is \(v = |\vec{v}_A| = \sqrt{v_{Ax}^2 + v_{Ay}^2}\). So, \(v^2 = (-a/t)^2 + v_1^2\). Rearranging for \(t\): \(t^2 = \frac{a^2}{v^2 - v_1^2}\), hence \(t = \frac{a}{\sqrt{v^2 - v_1^2}}\).
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)
1. 2 s
2. 5 s
3. 9 s
4. 8 s
View Answer
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
1. 15 m
2. 18 m
3. 30 m
4. 10 m
View Answer
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, \(S_3 = 2.5 + 2.5(5) = 15\text{ m}\).