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The acceleration - time graph of a particle is as shown in figure Initial velocity of particle is `10 m//s` Displacement of the particle (in m) at the end of 8 sec is ___ Round off to nearest integer. |
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Answer» Correct Answer - `0197` `v=u+at` `v_(1)=10+4xx4=26 m//s` `S_(1)=10xx4+1/2xx4xx16=40+32=72` m Now `a=8-t` `(dr)/(dt)=8-t` `underset26 overset rint dr = underset4overset t int(8-t)=dt` `v-26=[8t-t^(2)/2]_(4)^(t)` `v=26+(8t-t^(2)/2)-(32-16/2)` `v=26+8t-t^(2)/2 24` `v=2+8t-t^(2)/2` `(dx)/(dt)=2+8t-t^(2)/2` `x=[2t+4t^(2)-t^(3)/6]_(4)^(8)=(16+64xx4-8^(3)/6)-(8+64-64/6)=8+64xx3-74.6` `=200-74.6=125.33` Total distance `=72+125.33=197.33 approx 197]` |
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