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Uniform acceleration is a key concept in kinematics. It describes a situation where an object's velocity changes at a constant rate over time. This means the acceleration remains the same throughout the motion.
This simplifies calculations since you can use straightforward formulas to predict an object's future position or velocity.
In the exercise, the space vehicle accelerates uniformly from 65 m/s to 162 m/s over 10 seconds. The uniform acceleration (a) is calculated using the formula:

• \( a = \frac{v_f - v_0}{t} \)

where \( v_f \) is the final velocity, \( v_0 \) is the initial velocity, and \( t \) is the time period.
This calculation results in an acceleration of 9.7 m/s². Understanding uniform acceleration allows for predicting motion accurately within this condition, as it simplifies the complex real-world motions by breaking them down into calculable components.

Velocity refers to the speed of an object in a specified direction. It is a vector, which means it's expressed with both magnitude and direction.
In kinematics problems, velocity helps determine how quickly and in what direction an object is moving at any point in time.
For the space vehicle, we calculated velocities at specific times using the formula:

• \( v = v_0 + a \, t \)

Here, \( v \) represents the velocity at a specific time, \( v_0 \) is the initial velocity, and \( a \) is the uniform acceleration.
For example, at \( t = 2 \, \text{seconds} \), the velocity was 84.4 m/s, and at \( t = 6 \, \text{seconds} \), it was 122.2 m/s.
This illustrates how an object's velocity continuously adjusts under uniform acceleration, influencing how far it will travel over a certain period.

In kinematics, displacement is the change in position of an object, considering its initial and final locations. It differs from distance as displacement considers the shortest path, often involving direction.
Displacement is a vector quantity.
To calculate displacement with uniform acceleration, we used:

• \( s = ut + \frac{1}{2}a(t^2) \)

Here, \( s \) is displacement, \( u \) is the initial velocity for the time period, \( t \) is the time interval under consideration, and \( a \) is the acceleration.
In the exercise, the space vehicle traveled between \( t = 2 \, \text{seconds} \) and \( t = 6 \, \text{seconds} \) with an initial velocity of 84.4 m/s, calculated for this time period.
The resulting displacement was 415.2 meters, describing the vehicle's change in position concisely.