A vehicle travels half of the total distance with speed 2 m/s and the other half with speed 5 m/s, then its average speed is
Formula: For equal halves of distance, the average speed is the harmonic mean: \(v_{av} = \frac{2v_1v_2}{v_1+v_2}\). Putting values, \(v_{av} = \frac{2 \times 2 \times 5}{2+5} = \frac{20}{7}\text{ m/s}\).
A particle moving with uniform acceleration crosses two points A and B present in a straight line with speed 10 m/s and 20 m/s respectively, the speed of particle at mid-point of A and B will be
Formula: Under uniform acceleration, the midpoint velocity is \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}}\). Thus, \(v_{mid} = \sqrt{\frac{100 + 400}{2}} = 5\sqrt{10}\text{ m/s}\).
Two objects \(A\) & \(B\) are moving in a plane with velocities \(\vec{v}_A = (3\hat{i}+4\hat{j})\) m/s and \(\vec{v}_B = (7\hat{i}-3\hat{j})\) m/s respectively. The velocity of object \(A\) with respect to object \(B\) will be (in m/s):
Relative velocity is given by \(\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (3\hat{i}+4\hat{j}) - (7\hat{i}-3\hat{j}) = -4\hat{i}+7\hat{j}\) m/s. Therefore, option D is correct.
A car accelerates from rest at constant rate of \(2\text{ m/s}^2\) for some time after which it decelerates at a constant rate of \(3\text{ m/s}^2\) to come to rest. If total time taken for the motion is \(40\) seconds then maximum velocity achieved by the car during motion is
Using the relation \(v_{max} = \frac{\alpha \beta}{\alpha + \beta} t\), where \(\alpha = 2\) and \(\beta = 3\) are acceleration and deceleration rates respectively. Substituting the values: \(v_{max} = \frac{2 \times 3}{2 + 3} \times 40 = 48\text{ m/s}\).
A swimmer can swim with speed of \(8\text{ m/s}\) in still water. \(800\text{ m}\) wide river is flowing with speed of \(4\text{ m/s}\). Swimmer wants to cross the river in minimum time. Velocity of swimmer with respect to ground is (approximately)
For minimum crossing time, the swimmer must head perpendicular to the river bank. The net ground velocity is the vector sum of swimmer's velocity and river velocity: \(v_g = \sqrt{v_{\text{sw}}^2 + v_r^2} = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.94\text{ m/s} \approx 9\text{ m/s}\).
A particle is projected with a velocity \( \vec{v} = (3\hat{i} + 4\hat{j}) \text{ m/s} \) in the presence of uniform acceleration \( \vec{a} = (4\hat{i} – 3\hat{j}) \text{ m/s}^2 \). The path of the particle will be
Since the constant acceleration vector \( \vec{a} \) is not parallel or antiparallel to the initial velocity vector \( \vec{v} \) (here \( \vec{v} \cdot \vec{a} = 0 \)), the trajectory of the particle is parabolic.
In projectile motion if air resistance (or any of such force opposing motion) is taken into consideration, then
Air resistance acts opposite to the direction of motion, decreasing velocity. This causes deviation from the ideal symmetric parabolic trajectory and decreases both the range and maximum height. Thus, (1) and (2) are correct.
A particle is thrown vertically upward from ground. Its velocity at half of the height is \(10\text{ m/s}\), then maximum height attained by it (\(g = 10\text{ m/s}^2\))
Using third equation of motion: \(v^2 = u^2 - 2g(H/2) \Rightarrow 10^2 = u^2 - gH\). At max height, \(0 = u^2 - 2gH \Rightarrow u^2 = 2gH\). Thus, \(100 = 2gH - gH = gH \Rightarrow H = 10\text{ m}\).
A swimmer wants to cross the river in shortest possible time, The angle \(theta\) made by the swimmer with flow of river is
The time to cross a river is \(t = \frac{d}{v \sin \theta}\), where \(theta\) is the angle with the river flow. For \(t\) to be minimum, \(sin \theta\) must be maximum, which occurs at \(\theta = 90^\circ = \frac{\pi}{2}\).
Two particles A and B are projected from ground at an angle of 30° with the horizontal with velocity 20 m/s and 40 m/s respectively. The maximum height and time of flight are both greater for which particle?
Both maximum height \(H = \frac{u^2 \sin^2\theta}{2g}\) and time of flight \(T = \frac{2u \sin\theta}{g}\) are proportional to velocity \(u\). Since particle B has a larger velocity, both values are greater for B.