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In the context of kinematics, acceleration is the rate at which an object's velocity changes with time. It is a vector quantity, meaning it has both magnitude and direction. When a sprinter starts running, he goes from a state of rest to moving quickly in a short period. This rapid increase in speed is due to acceleration.

The formula for calculating acceleration is given by \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.

In the exercise, during the first 1.40 seconds, the sprinter accelerates from rest to a speed of 8.00 m/s. This means his acceleration can be calculated as \( a = \frac{8.00 \text{ m/s} - 0 \text{ m/s}}{1.40 \text{ s}} \).

• Initial velocity (\( v_i \)): 0 m/s
• Final velocity (\( v_f \)): 8.00 m/s
• Time interval (\( \Delta t \)): 1.40 s

Consequently, the sprinter's acceleration is 5.71 m/s², representing a maximum acceleration during this phase.

Velocity expresses both the speed and direction of an object's motion. Unlike speed, which is a scalar quantity only showing how fast an object is moving, velocity includes directional information.

In this scenario, we consider the sprinter's velocity as he undergoes constant acceleration. Initially, our sprinter accelerates to reach 8.00 m/s within 1.40 seconds from rest. We can use this to understand how much and how quickly his velocity is changing.

• Initial velocity (start at rest): 0 m/s
• Final velocity after 1.40 seconds: 8.00 m/s
• Top speed achieved after 7.02 seconds: 11.8 m/s

When he reaches his top speed of 11.8 m/s in 7.02 seconds from start, his average velocity during this period can also be used to understand the concept of average velocity, which is total displacement over total time. In kinematics, understanding changes in velocity over time is crucial as it directly ties with acceleration.

Distance travelled is a scalar quantity representing the total movement of an object regardless of direction. When dealing with constant acceleration, we often calculate distance using kinematic equations.

In this exercise, to determine how far the sprinter runs in the first 1.40 seconds, we apply the distance formula: \( d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \). Given that the initial velocity \( v_i \) is 0 m/s and acceleration \( a \) is 5.71 m/s², the formula simplifies, allowing us to find the distance traveled:

• Initial velocity (\( v_i \)): 0 m/s
• Time (\( t \)): 1.40 s
• Acceleration (\( a \)): 5.71 m/s²

Plugging these values in, we get \( d = 0 \cdot 1.40 + \frac{1}{2} \cdot 5.71 \cdot (1.40)^2 = 5.59 \text{ meters} \). This illustrates the path covered by the sprinter during constant acceleration.

Constant acceleration refers to a scenario where the acceleration of an object remains unchanged over time. In kinematic equations, this concept allows for simplifying calculations of speed, displacement, and time.

In the provided example, the sprinter exhibits constant acceleration in the initial 1.40 seconds. This is determined by the stable rate at which his velocity rises from 0 m/s to 8.00 m/s.

• Initially at rest: 0 m/s
• Final velocity in the initial phase: 8.00 m/s
• Time duration: 1.40 s

This consistent acceleration results in the sprinter maintaining a very precise and predictable motion, allowing us to easily compute the distance travelled using kinematic equations.
Knowing that acceleration stays consistent allows us to use the formula \( a = \frac{\Delta v}{\Delta t} \) without needing alterations through this phase. Utilizing constant acceleration simplifies many aspects of kinematic calculations and helps in modeling real-life motion scenarios efficiently.