QJEEM24Ph2ce1: JEE Main 2024 Physics - Class 11 JEEM24Ph2ce1 Solution: A particle is moving in one dimension (along axis) under the…

• Position (x): The particle's location on the x-axis at a given time.


• Velocity (v): The rate of change of position with respect to time

( $v = \frac{dx}{dt}$ ).

• Acceleration (a): The rate of change of velocity with respect to time

( $a = \frac{dv}{dt}$ ).

The particle's position is given by:

$$x(t)=-3t^3+18t^2+16t$$

We need to find the velocity when the acceleration is zero.

1. Find the Velocity (v) and Acceleration (a) Functions:

• Velocity:

$$v(t)=\frac{dx}{dt}=-9t^2+36t+16$$

• Acceleration:

$$a(t)=\frac{dv}{dt}=-18t+36$$

1. Find the Time (t) When Acceleration is Zero:

$$a(t)=0$$

$$-18t+36=0$$

$$t=2\text{ seconds}$$

1. Calculate the Velocity at t = 2 seconds:

$$v(2)=-9(2{)}^2+36(2)+16$$

$$v(2)=52\text{ m/s}$$

The velocity of the particle when its acceleration becomes zero is 52 m/s.