91Ó°ÊÓ

In kinematics, constant acceleration refers to a scenario where an object's acceleration remains unchanged over time. This means the change in velocity per unit of time is consistent, helping us predict and calculate motion more easily. In the exercise, the car accelerates with a steady rate as it moves along the ramp. Since its acceleration is constant, kinematic equations can simplify solving problems involving distance, velocity, and time. Constant acceleration is a common assumption to help solve problems in introductory physics, especially when analyzing straight-line motion.

Kinematic equations are a set of formulas used to describe and predict the motion of objects moving with constant acceleration. They relate variables such as initial velocity, final velocity, acceleration, time, and displacement.
Key kinematic equations include:

• \( v_f = v_i + at \) for final velocity.
• \( v_f^2 = v_i^2 + 2as \) for acceleration.
• \( s = v_i t + \frac{1}{2}at^2 \) for displacement.
• \( s = \frac{v_i + v_f}{2}t \) for average velocity.

In the ramp problem, we leverage the second equation to determine acceleration, given initial and final velocities and displacement.

Calculating acceleration involves using kinematic equations that relate it to changes in velocity and displacement. In the exercise, we apply the equation \( v_f^2 = v_i^2 + 2as \) since the car's initial speed is zero, streamlining the math. Simplifying gives us an equation solely with final speed and distance traveled. Plugging in the known values:
- Final velocity, \( v_f = 20 \, \text{m/s} \)- Initial velocity, \( v_i = 0 \, \text{m/s} \)- Distance, \( s = 120 \, \text{m} \)
The formula becomes \( a = \frac{v_f^2}{2s}\) and substituting numbers results in \( a = \frac{400}{240} = 1.67 \, \text{m/s}^2\).
This calculation confirms the uniform increase in speed as the car progresses along the ramp.

Calculating distance involves understanding how far an object travels over a period. In this exercise, it specifically pertains to traffic moving continuously at a constant speed while the car accelerates along the ramp. With a traffic velocity given as \( 20 \, \text{m/s} \) and using the time the car takes, which was calculated earlier to be around 12 seconds, we can find how far the traffic travels.

• Use the formula \( d = v \times t \).
• \( d = 20 \, \text{m/s} \times 12 \, \text{s} \).
• This results in \( d = 240 \, \text{m} \).

This shows us how the length covered by the car compares to that of vehicles already moving at freeway speed.