1
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-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)
2
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-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 :
x(t) = at + bt2 – ct3
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
3
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-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 ?
What is the magnitude of the acceleration at t = 1 ?
4
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
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)


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