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Acceleration measures how quickly an object's velocity changes. It's a vector, meaning it has both magnitude and direction. To calculate acceleration, we use the formula: , a= \( \frac{v_{f} - v_{i}}{t} \) Here, a stands for acceleration, v_f for final velocity, v_i for initial velocity, and t for the time taken. In the given problem, the car's initial velocity is 14 m/s, and it comes to a halt (so the final velocity is 0 m/s) in 4 seconds. Let's plug these values into our formula: a= \( \frac{0 \text{ m/s} - 14 \text{ m/s}}{4 \text{ s}} \) = \( \frac{-14}{4} \text{ m/s}^2 \), which simplifies to \( -3.5 \text{ m/s}^2 \). The negative sign is because the car is decelerating (slowing down).The car's acceleration when it stops is -3.5 m/s².

Velocity is the speed of something in a given direction. When the velocity of an object changes, it means there has been an acceleration or deceleration. The change in velocity can be calculated if we know the initial and final velocities and the time period over which the change occurs. To understand this, let's take another look at our original problem. We know the car starts with an initial velocity (v_i) of 20 m/s and we need to find out how long it will take to stop, given the same acceleration we calculated earlier (-3.5 m/s²). Using the formula for time: t= \( \frac{v_{f} - v_{i}}{a} \), we substitute \( v_{f} \) = 0 m/s (since the car stops), \( v_{i} \) = 20 m/s, and \( a \) = -3.5 m/s². t= \( \frac{0 \text{ m/s} - 20 \text{ m/s}}{-3.5 \text{ m/s}^2} \), = \( \frac{-20}{-3.5} \text{ s} \) which simplifies to 5.71 seconds. This application shows how knowing the change in velocity and the rate of acceleration can help us understand the time it takes for these changes to occur. Similarly, if we want to find out how long it takes to reduce the car's velocity from 20 m/s to 10 m/s, we use the same formula but change the final velocity to 10 m/s. t= \( \frac{10 \text{ m/s} - 20 \text{ m/s}}{-3.5 \text{ m/s}^2} \), which becomes \( \frac{-10}{-3.5} \text{ s} \), which simplifies to 2.86 seconds. This explains the idea of velocity change in terms of acceleration.

When an object experiences uniform acceleration, its velocity changes at a constant rate. This is a common scenario in many kinematics problems. Uniform acceleration allows us to use specific equations to describe the motion.One key equation used is:\[v_f =v_i + at\]where \( v_{f} \) is the final velocity, \( v_{i} \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.Since we've been working with this equation in our problem, we also know that there are other useful formulas such as:\[s = v_i t + \frac{1}{2}at^2 \]where \( s \) is the displacement.These equations can be combined and manipulated to find different unknowns depending on what information is provided in a problem.In our exercise, we've been using:\[ a = \frac{v_f - v_i}{t} \]Using these equations with uniform acceleration helps simplify calculations and enables us to solve various parts of a problem systematically. By understanding these fundamental equations, it's easier to tackle any kinematics problem that involves uniform acceleration.