Question

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 11
1.6 points save answer
an airplane that is flying level needs to accelerate from a speed of $2.00 \times 10^2$ m/s to a speed of $2.40 \times 10^2$ m/s while it flies a distance of 1.20 km. what must be the acceleration of the plane? see conversion factor from km to m.
$7.33$ m/s²
$4.44$ m/s²
$2.45$ m/s²
$5.78$ m/s²

Explanation:

Step1: Identify the kinematic equation

We use the third 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 \( a \), so rearrange the formula: \( a=\frac{v_f^2 - v_i^2}{2d} \).

Step2: Convert distance to meters

The distance \( d = 1.20\space km \). Since \( 1\space km = 1000\space m \), \( d = 1.20\times1000 = 1200\space m \).

Step3: Identify initial and final velocities

Initial velocity \( v_i = 2.00\times10^2\space m/s = 200\space m/s \), final velocity \( v_f = 2.40\times10^2\space m/s = 240\space m/s \).

Step4: Substitute values into the formula

Calculate \( v_f^2 - v_i^2 \): \( (240)^2 - (200)^2 = 57600 - 40000 = 17600\space m^2/s^2 \). Then, calculate \( 2d = 2\times1200 = 2400\space m \). Now, \( a=\frac{17600}{2400}\approx7.33\space m/s^2 \)? Wait, no, wait, let's recalculate: Wait, \( (240)^2 = 57600 \), \( (200)^2 = 40000 \), difference is \( 17600 \). Then \( 2d = 2\times1200 = 2400 \). Then \( 17600\div2400\approx7.33 \)? Wait, but let's check again. Wait, maybe I made a mistake. Wait, \( v_f = 240\space m/s \), \( v_i = 200\space m/s \), \( d = 1200\space m \). So \( a=\frac{(240)^2 - (200)^2}{2\times1200}=\frac{57600 - 40000}{2400}=\frac{17600}{2400}\approx7.33\space m/s^2 \). Wait, but let's check the options. The first option is \( 7.33\space m/s^2 \). Wait, but let's re-express the velocities: \( v_i = 2.00\times10^2 = 200 \), \( v_f = 2.40\times10^2 = 240 \). Distance \( d = 1.20\space km = 1200\space m \). So the calculation is correct. Wait, but let's check the arithmetic again: \( 240^2 = 57600 \), \( 200^2 = 40000 \), subtract: \( 17600 \). Divide by \( 2\times1200 = 2400 \). \( 17600\div2400 = 7.333...\approx7.33\space m/s^2 \). So the answer should be \( 7.33\space m/s^2 \).

Answer:

\( 7.33\space m/s^2 \) (corresponding to the first option: \( 7.33\space m/s^2 \))