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Question

Physics Question on laws of motion

A body of mass 2kg2 \,kg has an initial velocity of 3m/s3\, m/s along OEOE and it is subjected to a force of 4N4 \,N in a direction perpendicular to OEOE. The distance of body from OO after 4s4 \,s will be

A

12 m

B

20 m

C

8 m

D

48 m

Answer

20 m

Explanation

Solution

The acceleration of the body perpendicular to OEOE is a=Fm=42=2m/s2a=\frac{F}{m}=\frac{4}{2}=2\,m/{{s}^{2}} Displacement along OE, s1=vt=3×4=12m{{s}_{1}}=vt=3\times 4=12\,m Displacement perpendicular to OE s2=12at2{{s}_{2}}=\frac{1}{2}a{{t}^{2}} =12×2×(4)2=16m=\frac{1}{2}\times 2\times {{(4)}^{2}}=16\,m The resultant displacement s=s12+s22s=\sqrt{s_{1}^{2}+s_{2}^{2}} =144+256=\sqrt{144+256} =400=\sqrt{400} =20m=20\,m