A particle is moving such that its position coordinates (x, y) are: \((2\text{ m}, 3\text{ m})\text{ at time } t = 0,\) \((6\text{ m}, 7\text{ m})\text{ at time } t = 2\text{ s}\) and \((13\text{ m}, 14\text{ m})\text{ at time } t = 5\text{ s}\). Average velocity \((\vec{V}_{av})\text{ from } t = 0\text { to } t = 5\text{ s}\) is:
(2014)
1. \(\frac{1}{5}(13\hat{i}+14\hat{j})\)
2. \(\frac{7}{3}(\hat{i}+\hat{j})\)
3. \(2(\hat{i}+\hat{j})\)
4. \(\frac{11}{5}(\hat{i}+\hat{j})\)
View Answer
Average velocity is \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\vec{r}_f - \vec{r}_i}{t_f - t_i}\). Initial position at \(t_i = 0\text{ s}\) is \(\vec{r}_i = 2\hat{i} + 3\hat{j}\). Final position at \(t_f = 5\text{ s}\) is \(\vec{r}_f = 13\hat{i} + 14\hat{j}\). So, \(vec{V}_{av} = \frac{(13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j})}{5 - 0} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j})\) m/s.
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 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 \(\theta = 60^\circ\).
Consider the following two statements and tick the correct answer.
Statement A: In projectile motion, horizontal component of velocity of particle remains constant while its acceleration changes continuously.
Statement B: In projectile motion, particle moves such that its velocity and acceleration both changes continuously.
1. Statement A is correct while statement B is incorrect
2. Statement A is incorrect while statement B is correct
3. Both statement A and statement B are correct
4. Both statement A and statement B are incorrect
View Answer
In projectile motion, acceleration is constant throughout (equal to acceleration due to gravity \(\vec{g}\)). Thus, both statements are incorrect because they claim acceleration changes.
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, \(S_3 = 2.5 + 2.5(5) = 15\text{ m}\).