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A car moving with a velocity of 20 ms^(-1) is stopped in a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is |
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Answer» 640 m From NEWTON third equation, `v^(2)-u^(2)=2as` `(0)^(2)-(20)^(2)=2(a)xx40` `400=2xx40xxa` `a=(400)/(2xx40)` `a=5 ms^(-2)` In the SECOND condition, the velocity becomes twive i.e., `u^(1)=2u` Again from Newton.s third equation, we get `(0)^(2)-(2u)^(2)=2xx(5)xx s` `s=((2u)^(2))/(2xx5)` `s=(4u^(2))/(2xx5)` `s=(4xx20xx20)/(2xx5)` `s=(4xx20xx20)/(2xx5)` s = 160 m. |
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