AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

Potential energy,
U=2−20x+5x2J When particle gains whole of potential energy in the form of kinetic energy at x=−3m, then it will travel maximum distance till whole of its kinetic energy becomes zero.
i.e., ‌‌K=U=0
⇒‌‌2−20x+5x2=0
5x2−20x+2=0
=‌

20±√360

10

=‌

20±18.97

10

x=2±1.9
For maximum value of x, taking + ve sign
x=2+1.9=3.9
∴ Total distance =|−3|+39
=6.9m≃7m