Dear student, the distance covered by the car in the direction of east in 1h, with 60 Km/h=s1=speed×time=60×1=60 Km similarly the distance s2 covered in the direction of south in 12h with 60 Km/h=s2=speed×time=60×12=30 Km now the displacement from the initial position=S=s12+s22=602+302=304+1=305 Km Option 3 is correct Regards

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Dear student,

t h e d i s \tan c e c o v e r e d b y t h e c a r i n t h e d i r e c t i o n o f e a s t i n 1 h, w i t h 60 K m / h = s_{1} = s p e e d \times t i m e = 60 \times 1 = 60 K m . s i m i l a r l y t h e d i s \tan c e s_{2} c o v e r e d i n t h e d i r e c t i o n o f s o u t h i n \frac{1}{2} h w i t h 60 K m / h = s_{2} = s p e e d \times t i m e = 60 \times \frac{1}{2} = 30 K m . n o w t h e d i s p l a c e m e n t f r o m t h e i n i t i a l p o s i t i o n = S = \sqrt{{s_{1}}^{2} + {s_{2}}^{2}} = \sqrt{60^{2} + 30^{2}} = 30 \sqrt{4 + 1} = 30 \sqrt{5} K m .

Option 3 is correct

Regards

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