A) 12 m
B) 20 m
C) 8m
D) 48 m
Correct Answer: B
Solution :
The acceleration of the body perpendicular to OE is: \[\alpha =\frac{F}{m}=\frac{4}{2}=2\,m/{{s}^{2}}\] Displacement along\[OE,\] \[{{s}_{1}}=vt=3\times 4=12\,m\] Displacement perpendicular to\[OE\] \[{{s}_{2}}=\frac{1}{2}a{{t}^{2}}\] \[=\frac{1}{2}\times 2\times {{(4)}^{2}}=16\,m\] The resultant displacement \[s=\sqrt{s_{1}^{2}+s_{2}^{2}}\] \[=\sqrt{144+256}\] \[=\sqrt{400}\] \[=20\,m\]
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