Question

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description
show all you work, including units, on separate paper. follow the \problem solving method\.
this final exam is cumulative and covers material from the entire course.
instructions
from the list of choices, select the one best answer.
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question 9
1.6 points save answer
a car accelerates from 10.0 m/s to 30.0 m/s at a rate of 3.00 m/s². how far does the car travel while accelerating?
226 m
133 m
399 m
80.0 m

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \( v_f^2 = v_i^2 + 2ad \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance. We need to solve for \( d \), so rearrange the equation: \( d=\frac{v_f^2 - v_i^2}{2a} \).

Step2: Plug in the values

Given \( v_i = 10.0 \, \text{m/s} \), \( v_f = 30.0 \, \text{m/s} \), and \( a = 3.00 \, \text{m/s}^2 \). Substitute these into the formula:

$$ d=\frac{(30.0)^2 - (10.0)^2}{2\times3.00} $$

Step3: Calculate the numerator and denominator

First, calculate the numerator: \( (30.0)^2 - (10.0)^2 = 900 - 100 = 800 \). Then the denominator: \( 2\times3.00 = 6.00 \).

Step4: Solve for \( d \)

Divide the numerator by the denominator: \( d=\frac{800}{6.00}\approx133 \, \text{m} \).

Answer:

133 m (corresponding to the option "133 m")