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. \(9\text{ units}\)
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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. \(8.5\text{ units}\)
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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. \(30^\circ\)
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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. \(a/(v+v_1)\)
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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 \(t = \frac{a}{\sqrt{v^2 - v_1^2}}\).
The magnitude of vectors \(\vec{A},\vec{B}\) and \(\vec{C}\) are 3, 4 and 5 units respectively. If \(\vec{A} + \vec{B} = \vec{C}\) , the angle between \(\vec{A}\) and \(\vec{B}\) is:
(1988)
1. \(\pi/2\)
2. \(cos^{-1} (0.6)\)
3. \(tan^{-1} (7/5)\)
4. \(\pi/4\)
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Given \(|\vec{A}|=3, |\vec{B}|=4, |\vec{C}|=5\) and \(\vec{A} + \vec{B} = \vec{C}\). Squaring both sides: \(|\vec{A} + \vec{B}|^2 = |\vec{C}|^2\). \(|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|cos\theta = |\vec{C}|^2\). Substituting values: \(3^2 + 4^2 + 2(3)(4)cos\theta = 5^2\) which gives \(9 + 16 + 24cos\theta = 25\). Thus, \(24cos\theta = 0\) and \(\theta = \pi/2\).
The ‘x’ and ‘y’ coordinates of the particle at any time are \(x = 5t – 2t^2\) and \(y = 10t\), respectively, where ‘x’ and ‘y’ are in metres and ‘t’ in seconds. The acceleration of the particle at \(t = 2\text{ s}\) is:
(2017-Delhi)
1. \(5\text{ m/s}^2\)
2. \(-4\text{ m/s}^2\)
3. \(-8\text{ m/s}^2\)
4. \(0\)
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Given \(x = 5t - 2t^2\) and \(y = 10t\). Differentiating twice with respect to time to find acceleration. \(v_x = \frac{dx}{dt} = 5 - 4t\), \(a_x = \frac{dv_x}{dt} = -4\text{ m/s}^2\). \(v_y = \frac{dy}{dt} = 10\), \(a_y = \frac{dv_y}{dt} = 0\text{ m/s}^2\). The acceleration vector is \(\vec{a} = -4\hat{i}\). The x-component of acceleration is \(-4\text{ m/s}^2\), which is constant.