A) 12 m
B) 20 m
C) 8 m
D) 48 m
Correct Answer: B
Solution :
Displacement of body in 4 sec along OE \[{{s}_{x}}={{v}_{x}}t=3\times 4=12\ m\] Force along OF (perpendicular to OE) = 4 N \[\therefore \] \[{{a}_{y}}=\frac{F}{m}=\frac{4}{2}=2\ m/{{s}^{2}}\] Displacement of body in 4 sec along OF Þ \[{{s}_{y}}={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}\]\[=\frac{1}{2}\times 2\times {{(4)}^{2}}=16\ m\] [As \[{{u}_{y}}=0\]] \[\therefore \]Net displacement \[s=\sqrt{s_{x}^{2}+s_{y}^{2}}\ =\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=20\ m\]You need to login to perform this action.
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