A) 10
B) 15
C) 20
D) 30
Correct Answer: A
Solution :
Let the initial velocity = u and acceleration = a In first case \[{{s}_{1}}=u{{t}_{1}}+\frac{1}{2}at_{1}^{2}\] \[200=2u+2a\] \[(\because {{t}_{1}}=2s)\] \[u+a=100\] ...(i) In second case \[{{s}_{2}}=u{{t}_{2}}+\frac{1}{2}at_{2}^{2}\] \[420=6u+18a\] \[(\because {{t}_{2}}=2+4=6s)\] \[3a+u=70\] ...(ii) Solving Eqs. (i) and (ii), we get \[a=-15m/{{s}^{2}}\] and \[u=115m/s\] \[v=u+at\] \[=115-15\times 7=10m/s\]
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