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BHU PMT BHU PMT (Screening) Solved Paper-2006

  • question_answer
    A boy begins to walk eastward along a street in front of his house and the graph of his displacement from home is shown in the following figure. His average for the whole time interval is equal to:

    A)  8m/mm                              

    B)  6m/min

    C)  \[\frac{8}{3}m/\min \]                 

    D)  2 m/min

    Correct Answer: D

    Solution :

                     Key Idea: Average speed is the ratio of displacement to time taken. Displacement from 0 to\[5s=40\text{ }m\] Displacement from 5 to\[10\text{ }s=40\text{ }m\] Displacement from 0 to\[15\text{ }s=-\text{ }20m\] Displacement from 15 to\[20\text{ }s=0\] So, net displacement \[=40+40-20+0=60m\] Total time taken\[=5+5+15+5=30\text{ }mm\] Hence, average speed \[=\frac{displacement\text{ }(m)}{time(\min )}=\frac{60}{30}=2\text{ }m/min\]


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