91Ó°ÊÓ

In physics, kinematic equations are essential for solving problems related to motion. These equations describe how objects move in terms of position, velocity, and acceleration over time. They are particularly useful in scenarios where acceleration is constant. One of the fundamental kinematic equations used in this problem is:

• The equation: \( v^2 = u^2 + 2ad \) helps us find the acceleration needed when distance is known.
• Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the displacement.

In our given problem, we know:

• The final velocity \( v = 20 \, \text{m/s} \)
• The initial velocity \( u = 0 \, \text{m/s} \) because the car starts from rest
• The displacement \( d = 80 \, \text{m} \)

Using these values in the kinematic equation allows us to calculate the acceleration needed to achieve this movement over the specified distance.

Acceleration measures how quickly the velocity of an object changes. It is a key piece in understanding how forces cause changes in the motion of objects. The unit of acceleration is meters per second squared (\(\text{m/s}^2\)). To find the acceleration, we use the kinematic equation:

• \( v^2 = u^2 + 2ad \)

Given the final velocity \( v = 20 \, \text{m/s} \), initial velocity \( u = 0 \, \text{m/s} \), and displacement \( d = 80 \, \text{m} \), we plugged these values into the equation:\[(20)^2 = 0^2 + 2a \times 80\]This simplifies to:\[ 400 = 160a \]Solving for \( a \), we find:\[ a = \frac{400}{160} = 2.5 \, \text{m/s}^2 \]Thus, the car requires an acceleration of \( 2.5 \, \text{m/s}^2 \) to reach the desired velocity of \( 20 \, \text{m/s} \) over the distance of \( 80 \, \text{m} \).

Once we know the acceleration, we can determine the required force using Newton's Second Law of Motion, which connects force, mass, and acceleration. It is expressed as:

• \( F = ma \), where \( F \) is the force in newtons, \( m \) is the mass in kilograms, and \( a \) is the acceleration in meters per second squared.

For our car, we have:

• Mass \( m = 1300 \, \text{kg} \)
• Acceleration \( a = 2.5 \, \text{m/s}^2 \)

Using these values, we substitute into the formula:\[ F = 1300 \times 2.5 \]This calculation gives us:\[ F = 3250 \, \text{N} \]Hence, a force of \( 3250 \, \text{N} \) is necessary to accelerate the 1300-kg car from rest to a speed of \( 20 \, \text{m/s} \) over a distance of \( 80 \, \text{m} \). This highlights how forces interact with mass and motion, crucial aspects of Newtonian physics.