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A body travels 200 cm in the first two seconds and 220 cm in the next 4 seconds with deceleration. the velocity of the body at the end of the `7^(th)` second is |
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Answer» Correct Answer - `10cms^(-1)` Let `u` be initial velocity `&a ` be its acceleration Distance in first `2 sec =S_(1)=200cm` `rArr u(2)+(1)/(2)a(2)^(2)=200cm` `rArr u+a=100` ...`(i)` Distance in next `4 sec. =S_(2)=220cm` Distance in first `6 sec. =S_(1)+S_(2)=200+220cm` `rArru(6)+(1)/(2)a(6)^(2)=420` `rArr u+3a=70` ....`(ii)` From equations `(i) & (ii), ` we get `a=-15cm//s^(2),u=115 cm//s` Hence, velocity at the end of `7 sec.` from start `=u+7a` `=115+7(-15)=10cm//s`. |
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