91Ó°ÊÓ

Integration is a crucial concept in calculus, particularly when we need to find functions concerning their derivatives. In physics, integration helps translate an acceleration function into a velocity function, and subsequently a velocity function into a position function, describing a particle's motion over time.
To integrate a function, you essentially find the antiderivative, which is another function whose derivative would yield the original function. In this exercise, we began with the acceleration function \(a(t) = \sin t\).
By integrating this acceleration function with respect to time, we determined the velocity function \(v(t)\). When you integrate \(\sin t\) with respect to \(t\), you get \(-\cos t + C\), where \(C\) is the integration constant.

• The integration constant \(C\) is determined using initial conditions, such as a known velocity at a specific time.
• In our case, we used \(v(0) = 1\) to find \(C\), which helped finalize the velocity function to \(-\cos t + 2\).

This process shows how integration plays a pivotal role in connecting the physical concepts of motion, providing insights into velocity from a given acceleration.

Displacement is a vector quantity that refers to the change in position of an object. It is measured by the integral of the velocity function over a given time period.
Unlike distance, which is a scalar and considers the total ground covered, displacement specifically captures the net change in position over the time interval.
For the exercise at hand, once the velocity function \(v(t) = -\cos t + 2\) was established, finding displacement involves integrating this velocity function over the interval \([\pi/4, \pi/2]\).

• This is calculated through \(s(t) = \int_{\pi/4}^{\pi/2} (-\cos t + 2) \, dt\).
• By solving the integral, we obtained the displacement: \(\frac{\pi}{2} - 1 + \frac{\sqrt{2}}{2}\).

Thus, displacement not only tells us about where the particle started and ended but also considers the direction of travel.

The velocity function provides critical information about how an object's speed and direction change over time. In our scenario, velocity comes from integrating the given acceleration function.
The velocity function \(v(t) = -\cos t + 2\) was derived from the acceleration function \(a(t) = \sin t\) through integration.
This function tells us:

• How the speed of the particle changes over time.
• Whether the particle is speeding up or slowing down (depending on the sign of the velocity).

Understanding the velocity function helps determine if the particle changed direction during its motion, which is crucial for assessing how distance and displacement relate.
In this particular exercise, the velocity stayed positive over the interval \([\pi/4, \pi/2]\), showing no change in direction, confirming that displacement and distance were the same.
By examining the velocity function, you can effectively predict and model an object's movement under varying conditions, demonstrating its importance in calculus and physics.