A particle is moving in one dimension (along x-axis) under the action of a variable force. It's initial position was 16~m$16\mathrm{~m}$ right of origin. The variation of its position (x$x$) with time (t$t$) is given as x = -3t³ + 18t² + 16t$x = -3t^3 + 18t^2 + 16t$ where x$x$ is in m and t$t$ is in s. The velocity of the particle when its acceleration becomes zero is _______ m/s$\mathrm{m/s}$.

Related Formula

The relationship between position x$x$, velocity v$v$, and acceleration a$a$ is given by differentiation with respect to time t$t$:

Core Logic

Given position:

Differentiating once to find velocity v$v$:

Differentiating again to find acceleration a$a$:

Step 1: Determine Time when Acceleration is Zero

Set the acceleration to zero:

So, the acceleration becomes zero at t = 2 s$t = 2\text{ s}$.

Step 2: Calculate Velocity at this Time

Substitute t = 2 s$t = 2\text{ s}$ into the velocity equation:

Thus, the velocity is 8 m/s$8\text{ m/s}$.

Pattern Recognition

Sees: displacement function of degree 3 → acceleration is linear in time. The zero of a linear function of form At - B = 0$At - B = 0$ is easily calculated, and substituting back resolves to a basic quadratic.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Related Formula

Core Logic

Differentiating the position vector with respect to time gives the velocity vector:

Step 1: Evaluation at given time

Substitute t = 1 s$t = 1\text{ s}$ into the velocity expression:

Pattern Recognition

Constant term vectors (like 5 k$5\hat{k}$) drop out during differentiation. The resulting velocity only has a component along the j$\hat{j}$ direction, confirming motion purely parallel to the y-axis at that instant.

Chapter Mix

Class 11 Physics: Kinematics

Related Formula

Core Logic

Analyze the motion along the x-axis first to find the final x-velocity component (uₓ = 5 m/s$u_x = 5\text{ m/s}$, aₓ = 3 m/s²$a_x = 3\text{ m/s}^2$, x = 84 m$x = 84\text{ m}$):

Step 1: Compute time interval

Using the velocity relation along the x-axis:

Step 2: Evaluate y-axis velocity component

Since uy = 0$u_y = 0$ and ay = 2 m/s²$a_y = 2\text{ m/s}^2$, calculate vy$v_y$ at t = 6 s$t = 6\text{ s}$:

Step 3: Total net speed calculation

Comparing this with √(a)$\sqrt{a}$ yields a = 673$a = 673$.

Pattern Recognition

Splitting coordinates explicitly isolates vector calculation lines cleanly, optimizing equation selections relative to independent components.

Chapter Mix

Class 11 Physics: Kinematics

Related Formula

For a projectile fired horizontally from height h$h$ with speed u$u$:

Core Logic

To just clear step 4 and land on the tread of step 5, the flight trajectory must pass beyond the outer corner boundary vertex of the 4th$4^{\text{th}}$ step.

Therefore, net horizontal distance required to cross 4 complete steps is:

Similarly, net vertical fall matching the top of the 4th$4^{\text{th}}$ step level is:

Step 1: Determine Time of Flight

Using the vertical kinematic equation:

Step 2: Determine Velocity

Using the horizontal path equation:

Substituting range and time values:

Step 3: Extract x

Matching the parameter form u = √(x)$u = \sqrt{x}$, we find:

x = 2$x = 2$

Pattern Recognition

To clear the n$n$-th step, the projectile must safely pass the outer point of step (n-1)$(n-1)$. Treat the geometric coordinates of corners as bounding constraints (x = (n-1)w$x = (n-1)w$, y = (n-1)h$y = (n-1)h$) to configure kinematics instantaneously.

Chapter Mix

Class 11 Physics: Motion in a Plane