Jerk is defined as the rate of change of acceleration

Jerk is defined as the rate of change of acceleration: j = da dt = d2v dt2 = d3r dt3. Consider a particle moving in one dimension with constant jerk j. Determine the position x(t) of the particle in terms j and the initial values x0, v0, and a

The Correct Answer and Explanation is:

To determine the position x(t)x(t) of a particle moving in one dimension with constant jerk jj, we integrate the jerk successively:

Step 1: Jerk to Acceleration

Jerk is the time derivative of acceleration: j=dadtj = \frac{da}{dt}

Integrating once with respect to time tt: a(t)=jt+a0a(t) = jt + a_0

where a0a_0 is the initial acceleration at t=0t = 0.

Step 2: Acceleration to Velocity

Velocity is the integral of acceleration: v(t)=∫a(t) dt=∫(jt+a0) dt=12jt2+a0t+v0v(t) = \int a(t) \, dt = \int (jt + a_0) \, dt = \frac{1}{2}jt^2 + a_0t + v_0

where v0v_0 is the initial velocity.

Step 3: Velocity to Position

Position is the integral of velocity: x(t)=∫v(t) dt=∫(12jt2+a0t+v0)dtx(t) = \int v(t) \, dt = \int \left( \frac{1}{2}jt^2 + a_0t + v_0 \right) dt x(t)=16jt3+12a0t2+v0t+x0x(t) = \frac{1}{6}jt^3 + \frac{1}{2}a_0t^2 + v_0t + x_0

where x0x_0 is the initial position.


Final Answer:

x(t)=16jt3+12a0t2+v0t+x0\boxed{x(t) = \frac{1}{6}jt^3 + \frac{1}{2}a_0t^2 + v_0t + x_0}


Explanation:

In physics, jerk is the third derivative of position with respect to time and represents how acceleration changes over time. It is denoted by jj and defined as: j=dadt=d2vdt2=d3xdt3j = \frac{da}{dt} = \frac{d^2v}{dt^2} = \frac{d^3x}{dt^3}

If a particle moves with constant jerk jj, we can determine its position x(t)x(t) by integrating jerk successively.

First, integrating jerk yields the acceleration function. Since j=da/dtj = da/dt, integrating with respect to time gives: a(t)=jt+a0a(t) = jt + a_0

Here, a0a_0 is the constant of integration and represents the acceleration at t=0t = 0.

Next, integrating acceleration gives us the velocity: v(t)=∫a(t)dt=∫(jt+a0)dt=12jt2+a0t+v0v(t) = \int a(t) dt = \int (jt + a_0) dt = \frac{1}{2}jt^2 + a_0t + v_0

where v0v_0 is the initial velocity.

Finally, integrating velocity gives the position: x(t)=∫v(t)dt=∫(12jt2+a0t+v0)dt=16jt3+12a0t2+v0t+x0x(t) = \int v(t) dt = \int \left(\frac{1}{2}jt^2 + a_0t + v_0\right) dt = \frac{1}{6}jt^3 + \frac{1}{2}a_0t^2 + v_0t + x_0

with x0x_0 being the initial position.

Thus, the motion of a particle under constant jerk is described by a cubic function of time. Each term in this equation corresponds to the effect of jerk, acceleration, and velocity over time. The higher the jerk, the more rapidly acceleration (and thus velocity and position) changes. This function is useful in motion planning in robotics, animation, and vehicle dynamics, where smooth transitions in motion are critical.

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