1

### JEE Main 2017 (Online) 8th April Morning Slot

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity ?
A B C D ## Explanation

Given that,

acceleration (a) = $-$ C (constant)

$\therefore\,\,\,$ ${{dv} \over {dt}}$ = $-$ c

$\Rightarrow $$\,\,\, {{dv} \over {dx}} . {{dx} \over {dt}} = - c \Rightarrow$$\,\,\,$ $\upsilon$ ${{dv} \over {dx}}$ = $-$ c

$\Rightarrow $$\,\,\, \int {v\,dv} = - c\int {dx} \Rightarrow$$\,\,\,$ ${{{v^2}} \over 2}$ = $-$ cx + k

$\Rightarrow $$\,\,\, x = - {{{v^2}} \over {2c}} + {k \over c} From this equation we can say option (c) is the correct graph. 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2 . The car will catch up with the bus after a time of : A \sqrt {110} \,s B \sqrt {120} \,s C 10\,\,\sqrt 2 \,s D 15 s ## Explanation Acceleration of Car, aC = 4 m/s2 Acceleration of bus, aB = 2 m/s2 Initial distance between them, S = 200 m Acceleration of Car with respect to bus, aCB = aC - aB = 4 - 2 = 2 m/s2 As initially both are at rest so, uCB = 0 \therefore\,\,\, S = UCB \times t + {1 \over 2} aCB \times t2 \Rightarrow$$\,\,\,$ 200 = 0 + ${1 \over 2}$ $\times$ 2 $\times$ t2

$\Rightarrow $$\,\,\, t2 = 200 \Rightarrow$$\,\,\,$ t = 10$\sqrt 2$ sec.
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### JEE Main 2018 (Offline)

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.
A B C D ## Explanation

In option (A) you can see velocity versus time graph is a straight line with negative slope, so the acceleration is negative and constant. This motion is look like this. Initially velocity is maximum and as acceleration is negative so after travelling some distance velocity will will become zero and then velocity will increase in the negative direction. At option (B) you can see same situation happens so (A) and (B) represent same situation.

From diagram, Initially at t = 0 position h = 0 and after some time h become maximum at the end h again become zero. Option (D) represent this situation.

So, A, B and D are correct.

The distance - time graph for this case will be - But the given distance-time graph in the question does not match with this so option (C) will be wrong answer.
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### JEE Main 2018 (Online) 15th April Morning Slot

An automobile, travelling at $40\,$ km/h, can be stopped at a distance of $40\,$ m by applying brakes. If the same automobile is travelling at $80\,$ km/h, the minimum stopping distance, in metres, is (assume no skidding) :
A
$45\,$ m
B
$100\,$ m
C
$150\,$ m
D
$160\,$ m

## Explanation

In first case

u1 = 40 km / h
$\upsilon$1 = 0
s1 = 40 m

using, $\upsilon$2 $-$ $\mu$2 = 2 as

02 $-$ 402 = 2a $\times$ 40 . . . . . .(1)

In 2nd case

02 $-$ 802 = 2 as . . . . . .(2)

dividing (2) by (1) we get,

${s \over {40}}$ = ${{80{}^2} \over {{{40}^2}}}$

$\Rightarrow$ $\,\,\,\,$ S = ${{80 \times 80} \over {40}}$

$\Rightarrow$$\,\,\,\,$ S = 160 m