1
NEET 2016 Phase 2
+4
-1
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by
xP(t) = (at + bt2) and xQ(t) = (ft $$-$$ t2).

At what time do the cars have the same velocity ?
A
$${{a - f} \over {1 + b}}$$
B
$${{a + f} \over {2\left( {b - 1} \right)}}$$
C
$${{a + f} \over {2\left( {1 + b} \right)}}$$
D
$${{f - a} \over {2\left( {1 + b} \right)}}$$
2
NEET 2016 Phase 1
+4
-1
If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is
A
$${3 \over 2}A + {7 \over 3}B$$
B
$${A \over 2} + {B \over 3}$$
C
$${3 \over 2}A + 4B$$
D
$$3A + 7B$$
3
AIPMT 2015 Cancelled Paper
+4
-1
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $$v\left( x \right) = \beta {x^{ - 2n}}$$, where $$\beta$$ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by
A
$$- 2{\beta ^2}{x^{ - 2n + 1}}$$
B
$$- 2n{\beta ^2}{e^{ - 4n + 1}}$$
C
$$- 2n{\beta ^2}{x^{ - 2n - 1}}$$
D
$$- 2n{\beta ^2}{x^{ - 4n - 1}}$$
4
NEET 2013 (Karnataka)
+4
-1
The displacement 'x' (in meter) of a particle of mass 'm' (in kg) moving in one dimension under the action of a force, is related to time 't' (in sec) by t = $$\sqrt x + 3$$. The displacement of the particle when its velocity is zero, will be
A
4 m
B
0 m (zero)
C
6 m
D
2 m
EXAM MAP
Medical
NEET