1
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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)
A
$${{{b^2}\tau } \over {\sqrt 2 }}$$
B
$${{b^2}\tau }$$
C
$${{{b^2}\tau } \over 2}$$
D
$${{{b^2}\tau } \over 4}$$
2
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$$a + {{{b^2}} \over {c}}$$
B
$$a + {{{b^2}} \over {4c}}$$
C
$$a + {{{b^2}} \over {3c}}$$
D
$$a + {{{b^2}} \over {2c}}$$
3
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 ?
A
50
B
25
C
40
D
100
4
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time) JEE Main 2019 (Online) 8th April Evening Slot Physics - Motion in a Straight Line Question 89 English 1 JEE Main 2019 (Online) 8th April Evening Slot Physics - Motion in a Straight Line Question 89 English 2 JEE Main 2019 (Online) 8th April Evening Slot Physics - Motion in a Straight Line Question 89 English 3 JEE Main 2019 (Online) 8th April Evening Slot Physics - Motion in a Straight Line Question 89 English 4
A
(B), (C)
B
(A)
C
(A), (B), (C)
D
(A), (B), (D)
JEE Main Subjects
EXAM MAP