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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

  • question_answer
    An astronaut has left the international space station to test a new space scooter. His partner measures the following velocity changes which take place in 10 s interval. Find the magnitude and direction of average acceleration. At the beginning of 10 s interval the astronaut is moving towards positive x-axis at 10 m/s and at the end of, 10 s he is moving towards negative x-axis at 5 m/s.

    A)  15 m/s2, along positive x-axis

    B)  15 m/s2, along negative x-axis

    C)  1.5 m/s2, along negative x-axis

    D)  1.5 m/s2, along positive x-axis

    Correct Answer: C

    Solution :

    Here direction of velocity is changing after some time interval, ie, initial velocity and acceleration are in opposite sign \[{{a}_{av}}=\frac{({{\overrightarrow{v}}_{2}}-{{\overrightarrow{v}}_{1}})}{10}=\frac{(-5\hat{i}-10\hat{i})}{10}\]\[=-1.5\hat{i}\,m\text{/}{{s}^{2}}\] \[\Rightarrow \]\[1.5\,\text{m/}{{\text{s}}^{2}}\]along negative x-axis.


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