A particle is moving along positive \(x\text-\)axis. Its position varies as \(x = t^3-3t^2+12t+20,\) where \(x\) is in meters and \(t\) is in seconds. The velocity of the particle when its acceleration zero is:
1. \(1~\text{m/s}\)
2. \(3~\text{m/s}\)
3. \(6~\text{m/s}\)
4. \(9~\text{m/s}\)
Two forces and are acting on a particle.
The resultant force acting on particle is:
(A)
(B)
(C)
(D)
and , then angle between vectors A and B is:
(1)
(2)
(3)
(4)
If vectors \(\overrightarrow{{A}}=\cos \omega t \hat{{i}}+\sin \omega t \hat{j}\) and \(\overrightarrow{{B}}=\cos \left(\frac{\omega t}{2}\right)\hat{{i}}+\sin \left(\frac{\omega t}{2}\right) \hat{j}\) are functions of time. Then, at what value of \(t\) are they orthogonal to one another?
1. \(t = \frac{\pi}{4\omega}\)
2. \(t = \frac{\pi}{2\omega}\)
3. \(t = \frac{\pi}{\omega}\)
4. \(t = 0\)
Six vectors through have the magnitudes and directions indicated in the figure. Which of the following statements is true?
1.
2.
3.
4.
\(\overrightarrow{A}\) and \(\overrightarrow B\) are two vectors and \(\theta\) is the angle between them. If \(\left|\overrightarrow A\times \overrightarrow B\right|= \sqrt{3}\left(\overrightarrow A\cdot \overrightarrow B\right),\) then the value of \(\theta\) will be:
1. | \(60^{\circ}\) | 2. | \(45^{\circ}\) |
3. | \(30^{\circ}\) | 4. | \(90^{\circ}\) |
If a curve is governed by the equation y = sinx, then the area enclosed by the curve and x-axis between x = 0 and x = is (shaded region):
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units
The acceleration of a particle starting from rest varies with time according to relation, . The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)
1.
2.
3.
4.
The displacement of the particle is zero at \(t=0\) and at \(t=t\) it is \(x\). It starts moving in the \(x\)-direction with a velocity that varies as \(v = k \sqrt{x}\), where \(k\) is constant. The velocity will: (Here, \(v=\frac{dx}{dt}\))
1. | vary with time. |
2. | be independent of time. |
3. | be inversely proportional to time. |
4. | be inversely proportional to acceleration. |
The acceleration of a particle is given as \(a= 3x^2\).
At \(t=0,v=0\) and \(x=0\). It can then be concluded that the velocity at \(t=2~\text{s}\) will be: (Here, \(a=v\frac{dv}{dx}\))
1. \(0.05~\text{m/s}\)
2. \(0.5~\text{m/s}\)
3. \(5~\text{m/s}\)
4. \(50~\text{m/s}\)