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# A force of $(6 \hat i+8 \hat j) N$ acted on a body of mass 10 kg.The displacement after 10 sec. if it starts from rest will be

a) 50 m along $tan ^{-1} 4/3$ with x axis b) 70 m along $tan ^{-1} 3/4$ with x axis c) 10 m along $tan ^{-1} 4/3$ with x axis d) None

Answer: 50m along $\tan^{-1} \large\frac{4}{3}$ with x-axis
Acceleration $a=\large\frac{\hat F}{m}=\large\frac{6 \hat i+8 \hat j}{10}$
Since we know the acceleration, we can use the Kinematics equation $\overrightarrow s=\overrightarrow u t+\large\frac{1}{2} $$\overrightarrow a t^2 to get the displacement. \quad=0+\large\frac{1}{2}\bigg(\frac{6 \hat i+8 \hat j}{10}\bigg)$$10 ^2 $$=30 \hat i+40 \hat j Magnitude = \sqrt {30 ^2 +40 ^2}= 50\;m The direction is along an angle of \tan ^{-1}\large\frac{40}{30}$$=\tan ^{-1} \;\large\frac{4}{3}$ with x-axis
edited Aug 20, 2014