A)
B)
C)
D)
Correct Answer: D
Solution :
[d] \[{{x}_{1}}=\frac{1}{2}a{{t}^{2}}\] and \[{{x}_{2}}=vt\] |
\[\therefore {{x}_{1}}-{{x}_{2}}=\frac{1}{2}a{{t}^{2}}-vt\] |
\[\Rightarrow \,{{x}_{12}}=\frac{1}{2}a{{t}^{2}}-vt\] |
At \[=x,\,{{x}_{12}}=0\] and at any time \[\frac{{{d}^{2}}{{x}_{12}}}{d{{t}^{2}}}>0\] |
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