The velocity of a particle traveling along a straight line is v=(3 t^2-6 t) ft / s, where t is i

step 1 Well, position is a function of time can be written is integral of v times d t. So this integral

The velocity of a particle traveling along a straight line...

The velocity of a particle traveling along a straight line is $v=\left(3 t^{2}-6 t\right) \mathrm{ft} / \mathrm{s},$ where $t$ is in seconds. If $s=4 \mathrm{ft}$ when $t=0,$ determine the position of the particle when $t=4$ s. What is the total distance traveled during the time interval $t=0$ to $t=4 \mathrm{s} ?$ Also, what is the acceleration when $t=2 \mathrm{s} ?$

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Key Concepts

Displacement vs. Total Distance Traveled

Displacement is the net change in position and can be found by evaluating the position function at the endpoints of the time interval. Total distance traveled, however, accounts for the entire path taken, not just the net change. It is computed by integrating the absolute value of the velocity function over the time interval, ensuring that reversals in direction are correctly accounted for.

Acceleration as the Derivative of Velocity

Acceleration is defined as the rate of change of velocity with respect to time. In kinematics, it is obtained by differentiating the velocity function. Evaluating the acceleration function at a specific time gives the instantaneous acceleration at that moment.

Integration of Velocity

This concept involves finding the position function by integrating the velocity function with respect to time. When the velocity of a particle is known, integrating it yields the displacement function plus a constant. The constant is then determined using the initial condition, which specifies the position of the particle at a particular time.

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