The position vector of a particle \(\vec{R}\) as a function of time is given by: \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\) Where R is in metres, t is in seconds and \(\hat{i}\) and \(\hat{j}\) denote unit vectors along x and y-direction, respectively. Which one of the following statements is wrong for the motion of particle?
(2015)
1. Path of the particle is a circle of radius 4 metre
2. Acceleration vectors is along \(-vec{R}\)
3. Magnitude of acceleration vector is \(\frac{V^2}{R}\) where V is the velocity of particle.
4. Magnitude of the velocity of particle is 8 metre/second
View Answer
From \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\), \(x=4sin(2\pi t)\) and \(y=4cos(2\pi t)\). \(x^2+y^2=16\) implies a circle of radius 4m. \(\vec{V} = 8\pi cos(2\pi t)\hat{i} - 8\pi sin(2\pi t)\hat{j}\). \(|\vec{V}| = 8\pi\text{ m/s}\). \(\vec{a} = -16\pi^2sin(2\pi t)\hat{i} - 16\pi^2cos(2\pi t)\hat{j} = -4\pi^2 \vec{R}\). So \(\vec{a}\) is along \(-\vec{R}\). Also, \(|\vec{a}| = 16\pi^2\) and \(\frac{V^2}{R} = \frac{(8\pi)^2}{4} = 16\pi^2\). Therefore, (a), (b), (c) are correct. (d) is wrong because \(|\vec{V}| = 8\pi\text{ m/s}\), not 8 m/s.
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.
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.
1. \(3\sqrt{2} \text{ms}^{-1}, 30^\circ\)
2. \(3\sqrt{2} \text{ms}^{-1}, 45^\circ\)
3. \(4\sqrt{2} \text{ms}^{-1}, 45^\circ\)
4. \(4\sqrt{2} \text{ms}^{-1}, 60^\circ\)
View Answer
Velocity \(\vec{v}(t) = \frac{d\vec{r}}{dt} = 4\hat{i} + 4\that{j}\). At \(t = 1 \text{s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2} \text{m/s}\). The angle with the x-axis is \(tan\theta = \frac{v_y}{v_x} = \frac{4}{4} = 1 ⇒
\theta = 45^\circ\).
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 particle is moving in x-y plane such that its x and y coordinates changes with time according to relation, \(x = 3t^2\) & \(y = 5t\) (here x & y are in m & t is in s). Speed of the particle at \(t = 2\) s, will be
1. 17 \(\text{m s}^{-1}\)
2. \(\sqrt{34}\text{ m s}^{-1}\)
3. 13 \(\text{m s}^{-1}\)
4. 11 \(\text{m s}^{-1}\)
View Answer
The velocity components are \(v_x = \frac{dx}{dt} = 6t\) and \(v_y = \frac{dy}{dt} = 5\). At \(t = 2\) s, \(v_x = 12\text{ m/s}\) and \(v_y = 5\text{ m/s}\). Speed is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 5^2} = 13\text{ m/s}\).
Given below are two statements one is labelled as assertion (A) and reason (R)
Assertion: The length of actual path travelled by a body in given time interval is always equal to displacement.
Reason: If displacement is zero, then body is either at rest or it has returned to initial position.
Choose the correct option.
1. Assertion is true but Reason is false
2. Assertion is false but Reason is true
3. Both Assertion and Reason are true and Reason is correct explanation of Assertion
4. Both Assertion and Reason are true and Reason is not the correct explanation of Assertion
View Answer
Distance (actual path length) is greater than or equal to displacement magnitude, so the assertion is false. If displacement is zero, the body either remained at rest or returned to its starting point, making the reason true.
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, \(S_3 = 2.5 + 2.5(5) = 15\text{ m}\).