The potential energy (in SI units) of a particle of mass 2kg in a conservative field is U=6x−8y. If the initial velocity of the particle is →u=−1.5^i+2^jm/s, then the total distance travelled by the particle in first two seconds is
A
10m
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B
12m
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C
15m
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D
18m
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Solution
The correct option is C15m →F=−∂U∂x^i−∂U∂y^j=−[6^i]+[8]^j=−6^i+8^j ∴→a=−3^i+4^j m/s2 has same direction as that of →u=−3^i+4^j2 m/s
So, ∣∣→a∣∣=5 m/s2,∣∣→u∣∣=52 m/s
Since →u and →a are in same direction, particle will move along a straight line. S=ut+12at2
Given t=2s ∴S=52×2+12×5×22=5+10=15m