In position-time graphs, the slope represents speed, and the y-intercept at a given time represents distance.
(A), (C) and (E) only
(A) and (E) only
(B) and (E) only
(A), (C) and (D) only
The slope of A is given as:
\( \text{slope}_A = V_A \)
The slope of B is given as:
\( \text{slope}_B = V_B \)
Since:
\( \text{slope}_B > \text{slope}_A \)
It follows that:
\( V_B > V_A \)
As velocity (\( V \)) is inversely related to time (\( t \)) for a given distance:
\( t_B < t_A \)
The final answer is: (A) and (E) only.
A person moved from A to B on a circular path as shown in figure If the distance travelled by him is 60 m, then the magnitude of displacement would be Given ( Cos 135° = -0.7)
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: