The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
The distance covered by an object is given by the area under the velocity-time graph.
Here, the graph shows a combination of trapezoidal and rectangular areas.
The total area under the graph between \( t = 0 \) and \( t = 4 \) represents the distance traveled.
By calculating the area from the graph: \[ {Distance} = {Area under the graph} = 13 \, {m} \]
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: