Question

1. a car exiting the highway begins with a speed of 25 m/s and travels down a 100m - long exit ramp with a deceleration of 3m/s². what is the cars velocity at the end of the exit ramp?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}+2ax$, where $v$ is the final velocity, $v_{0}$ is the initial velocity, $a$ is the acceleration, and $x$ is the displacement.

Step2: Substitute the given values

Given $v_{0} = 25\ m/s$, $a=- 3\ m/s^{2}$ (negative because it's deceleration), and $x = 100\ m$.
Substitute into the equation: $v^{2}=(25)^{2}+2\times(-3)\times100$.
First, calculate $(25)^{2}=625$ and $2\times(-3)\times100=-600$.
Then $v^{2}=625 - 600=25$.

Step3: Solve for $v$

Take the square - root of both sides: $v=\sqrt{25}=5\ m/s$.

Answer:

$5\ m/s$