91Ó°ÊÓ

In physics, constant acceleration means that an object's velocity changes at a consistent rate over time. This condition is common in uniformly accelerated motion, like the plane in the exercise. When acceleration is constant, the kinematic equations become very useful. These equations help in predicting the future position and velocity of an object under constant acceleration. In our example, the plane undergoes constant acceleration as it speeds up along the runway. The acceleration doesn't change, allowing us to use kinematic equations to solve for the distance traveled or the final velocity.

Initial velocity is the speed of an object when the observation begins. This is usually denoted as 'u'. In the exercise, the plane starts from rest, meaning its initial velocity is zero. Initial velocity is crucial for calculating other variables such as acceleration, time, and final velocity. Knowing the initial velocity helps set up the kinematic equations appropriately. For example, because the plane starts from rest, the kinematic equation simplifies, such sections of the equation involving initial velocity can be omitted or simplified.

Final velocity is the speed of an object at the end of a given period of time. This is denoted as 'v'. In the exercise, we were required to find the final velocity of the plane as it takes off. We achieved this by first determining the acceleration using the given distance and time, and then plugging the acceleration value into another kinematic equation. Final velocity is a crucial part of understanding motion because it gives insight into how fast an object is moving at the end of its observational period. The formula we used is: \ \ \( v = u + at \)
Here, 'u' is the initial velocity, 'a' is constant acceleration, and 't' is the time taken.

The distance traveled, denoted as 's', is the total length of the path taken by an object during its motion. In our exercise, the plane covers 450 meters before takeoff. The distance is a key input for the kinematic equations, allowing us to solve for other variables like acceleration or time. In our example, knowing the distance along with initial velocity and time helped us find the plane's acceleration by using the equation: \ \( s = ut + \frac{1}{ 2}at^2 \)
Since the plane started from rest (\( u = 0 \)), the equation simplified to: \( s = \frac{1 }{2 }a t^2 \). This easier form made it possible to solve for acceleration first, and then subsequently use that result to find the final velocity.

Time, denoted as 't', is the duration over which motion occurs. In our example, the plane takes 15 seconds to travel 450 meters. Time is a fundamental aspect of kinematics because it ties directly to velocity and acceleration. Measuring time helps us track how quickly changes in motion happen. In the exercise, knowing the time allowed us to substitute it into kinematic equations. This way, we calculated both the acceleration and final velocity of the plane. The kinematic relationship helping us here was: \( s = ut + \frac{1}{2 }at^2 \), which reorganizes into: \( a = \frac{2s}{ t^2} \). With given distance and time, this formula allowed us to determine acceleration efficiently.