A) 20
B) 30
C) 50
D) 80
Correct Answer: C
Solution :
Distance travelled in \[{{t}^{th}}\] second of uniformly accelerated motion is \[{{s}_{t}}=u+\frac{a}{2}(2t-1)\] - (i) Distance travelled in \[{{(t+1)}^{th}}\] second can be written as \[{{s}_{t+1}}=u+\frac{a}{2}[2(t+1)-1]\] Or \[{{s}_{t+1}}=u+\frac{a}{2}(2t+1)\] \[\because {{s}_{1}}+{{s}_{t+1}}=100cm\] (given) \[\therefore u+\frac{a}{2}(2t-1)+u+\frac{a}{2}(2t+1)=100\] Or \[u+at-\frac{a}{2}+u+at+\frac{a}{2}=100\] Or \[2u+2at=100\] or \[u+at=50\] \[\therefore v=50\,\,cm\,\,{{s}^{-1}}\]You need to login to perform this action.
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