A) 30 sec
B) 15 sec
C) 10 sec
D) 5 sec
Correct Answer: A
Solution :
If a body starts from rest with acceleration a and then retards with retardation P and comes to rest. The total time taken for this journey is t and distance covered is S then \[S=\frac{1}{2}\frac{\alpha \beta {{t}^{2}}}{(\alpha +\beta )}\,\,=\,\,\frac{1}{2}\frac{5\times 10}{(5+10)}\,\times \,\,{{t}^{2}}\] \[\Rightarrow \,\,1500\,\,=\,\,\frac{1}{2}\frac{5\times 10}{(5+10)}\,\times \,\,{{t}^{2}}\] \[t\,\,=\,\,30\,sec\]
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