JEE MAIN - Physics (2018 (Offline) - No. 22)
All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.
_en_22_1.png)
_en_22_2.png)
_en_22_3.png)
_en_22_4.png)
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.
_en_22_5.png)
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 -
_en_22_6.png)
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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