NEET 2016 Phase 2 | Motion in a Straight Line Question 14 | Physics

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by
x_P(t) = (at + bt²) and x_Q(t) = (ft $$-$$ t²).

At what time do the cars have the same velocity ?

$${{a - f} \over {1 + b}}$$

$${{a + f} \over {2\left( {b - 1} \right)}}$$

$${{a + f} \over {2\left( {1 + b} \right)}}$$

$${{f - a} \over {2\left( {1 + b} \right)}}$$

NEET 2016 Phase 1

MCQ (Single Correct Answer)

-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

$${3 \over 2}A + {7 \over 3}B$$

$${A \over 2} + {B \over 3}$$

$${3 \over 2}A + 4B$$

$$3A + 7B$$

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)

-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

$$ - 2{\beta ^2}{x^{ - 2n + 1}}$$

$$ - 2n{\beta ^2}{e^{ - 4n + 1}}$$

$$ - 2n{\beta ^2}{x^{ - 2n - 1}}$$

$$ - 2n{\beta ^2}{x^{ - 4n - 1}}$$

NEET 2013 (Karnataka)

MCQ (Single Correct Answer)

-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

4 m

0 m (zero)

6 m

2 m

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NEET Subjects

Biology

Zoology

Human Health and Diseases Body Fluids and Its Circulation Locomotion and Movement Neural Control and Coordination Reproduction in Organisms Reproductive Health Structural Organisation in Animals Digestion and Absorption Excretory Products and Their Elimination Chemical Coordination and Integration Human Reproduction Animal Kingdom Breathing and Exchange of Gases Evolution