Question

For a particle moving under the action of a variable force, the kinetic energy ($k$) versus position ($x$) graph is given, then :

• A. At $A$, the particle is decelerating
• B. At $B$, the particle decelerating
• C. At $C$, the particle has maximum velocity
• D. At $D$, the particle has maximum acceleration

Answer 1

The correct option is D At $D$, the particle has maximum acceleration

$K.E=\frac{1}{2}m{v}^2$

$\frac{dK.E}{dt}=mv\frac{dv}{dt}=mva$

At D,$\frac{dK.E}{dt}=$maximum

So accelerating of particle is maximum at D.