JEE Main 2020 (Online) 2nd September Morning Slot | Motion in a Straight Line Question 77 | Physics

Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/ hour. Speed (in ms^–1) of this person as observed from train B will be close to :
(take the distance between the tracks as negligible)

30.5 ms^–1

29.5 ms^–1

31.5 ms^–1

28.5 ms^–1

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

-1

A particle is moving with speed v = b$$\sqrt x $$ along positive x-axis. Calculate the speed of the particle at time t = $$\tau $$(assume that the particle is at origin t = 0)

$${{{b^2}\tau } \over {\sqrt 2 }}$$

$${{b^2}\tau }$$

$${{{b^2}\tau } \over 2}$$

$${{{b^2}\tau } \over 4}$$

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

The position vector of a particle changes with time according to the relation $$\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat j$$
What is the magnitude of the acceleration at t = 1 ?

100

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

The position of a particle as a function of time t, is given by
x(t) = at + bt² – ct³
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

$$a + {{{b^2}} \over {c}}$$

$$a + {{{b^2}} \over {4c}}$$

$$a + {{{b^2}} \over {3c}}$$

$$a + {{{b^2}} \over {2c}}$$

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