Q. No.The \(x\) and \(y\) coordinates of the particle at any time are \(x=5 t-2 t^2\) and \({y}=10{t}\) respectively, where \(x\) and \(y\) are in meters and \(\mathrm{t}\) in seconds. The acceleration of the particle at \(\mathrm{t}=2\) s is:
1.
\(5\hat{i}~\text{m/s}^2\)
2.
\(-4\hat{i}~\text{m/s}^2\)
3.
\(-8\hat{j}~\text{m/s}^2\)
4.
\(0\)
| 1. | The velocity and acceleration both are parallel to \(\vec{r }.\) |
| 2. | The velocity is perpendicular to \(\vec{r }\) and acceleration is directed towards to origin. |
| 3. | The velocity is parallel to \(\vec{r }\) and acceleration is directed away from the origin. |
| 4. | The velocity and acceleration both are perpendicular to \(\vec{r}.\) |
| 1. | \(0.15\) m/s2 | 2. | \(0.18\) m/s2 |
| 3. | \(0.2\) m/s2 | 4. | \(0.1\) m/s2 |
A ship \(A\) is moving westward with a speed of \(10~\text{kmph}\) and a ship \(B,\) \(100 ~\text{km}\) south of \(A,\) is moving northward with a speed of \(10~\text{kmph}.\) The time after which the distance between them becomes the shortest is:
1. \(0~\text{h}\)
2. \(5~\text{h}\)
3. \(5\sqrt{2}~\text{h}\)
4. \(10\sqrt{2}~\text{h}\)
Two particles \({A}\) and \({B}\), move with constant velocities \(\vec{v}_1\) and \(\vec{v}_2\) respectively. At the initial moment, their position vectors are \(\vec{r}_1\) and \(\vec r_2\) respectively. The conditions for particles \({A}\) and \({B}\) for their collision will be:
| 1. | \(\dfrac{\vec{r}_1-\vec{r}_2}{\left|\vec{r}_1-\vec{r}_2\right|}=\dfrac{\vec{v}_2-\vec{v}_1}{\left|\vec{v}_2-\vec{v}_1\right|}\) |
| 2. | \(\vec{r}_1 \cdot \vec{v}_1=\vec{r}_2 \cdot \vec{v}_2\) |
| 3. | \(\vec{r}_1 \times \vec{v}_1=\vec{r}_2 \times \vec{v}_2\) |
| 4. | \(\vec{r}_1-\vec{r}_2=\vec{v}_1-\vec{v}_2\) |
\(\vec{{R}}=4 \sin (2 \pi {t}) \hat{i}+4 \cos (2 \pi {t}) \hat{j},\)
where \(R\) is in metres, \(t\) is in seconds and \({\hat{i},\hat{j}}\) denotes unit vectors along \({x}\) and \({y}\text-\)directions, respectively. Which one of the following statements is wrong for the motion of the particle?
| 1. | Acceleration is along \((\text{-}\vec R )\). |
| 2. | Magnitude of the acceleration vector is \(\frac{v^2}{R}\), where \(v\) is the velocity of the particle. |
| 3. | Magnitude of the velocity of the particle is \(8\) m/s. |
| 4. | Path of the particle is a circle of radius \(4\) m. |
A projectile is fired from the surface of the earth with a velocity of \(5~\text{m/s}\) and at an angle \(\theta\) with the horizontal. Another projectile fired from another planet with a velocity of \(3~\text{m/s}\) at the same angle follows a trajectory that is identical to the trajectory of the projectile fired from the Earth. The value of the acceleration due to gravity on the other planet is: (given \(g=9.8~\text{m/s}^2\) )
1. \(3.5~\text{m/s}^2\)
2. \(5.9~\text{m/s}^2\)
3. \(16.3~\text{m/s}^2\)
4. \(110.8~\text{m/s}^2\)
The velocity of a projectile at the initial point \(A\) is \(2\hat i+3\hat j~\text{m/s}.\) Its velocity (in m/s) at the point \(B\) is:
| 1. | \(-2\hat i+3\hat j~\) | 2. | \(2\hat i-3\hat j~\) |
| 3. | \(2\hat i+3\hat j~\) | 4. | \(-2\hat i-3\hat j~\) |
The speed of a swimmer in still water is \(20~\text{m/s}.\) The speed of river water is \(10~\text{m/s}\) and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes with respect to the north is given by:
| 1. | \(45^{\circ}\) west of north | 2. | \(30^{\circ}\) west of north |
| 3. | \(0^{\circ}\) west of north | 4. | \(60^{\circ}\) west of north |
Two particles \(A\) and \(B\) are moving in a uniform circular motion in concentric circles of radii \(r_A\) and \(r_B\) with speeds \(v_A\) and \(v_B\) respectively. Their time periods of rotation are the same. The ratio of the angular speed of \(A\) to that of \(B\) will be:
| 1. | \( 1: 1 \) | 2. | \(r_A: r_B \) |
| 3. | \(v_A: v_B \) | 4. | \(r_B: r_A\) |
© 2025 GoodEd Technologies Pvt. Ltd.