QuestionJune 13, 2025

A 2500 kg car initially at rest rolls down a frictionless hill that is inclined at 37.2^circ If the hill is 30.0 m long what is the velocity of the car as it reaches the base of the hill? a=5.93m/s^2 v_(f)=[?]m/s

A 2500 kg car initially at rest rolls down a frictionless hill that is inclined at 37.2^circ If the hill is 30.0 m long what is the velocity of the car as it reaches the base of the hill? a=5.93m/s^2 v_(f)=[?]m/s
A 2500 kg car initially at rest rolls down a frictionless hill that is inclined at 37.2^circ If the hill is 30.0 m long what is the velocity of the car as it reaches the base of the hill? a=5.93m/s^2 v_(f)=[?]m/s

Solution
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Answer

v_f = 18.8 \, \text{m/s} Explanation 1. Calculate the acceleration Use a = g \cdot \sin(\theta) where g = 9.81 \, \text{m/s}^2 and \theta = 37.2^\circ. Thus, a = 9.81 \cdot \sin(37.2^\circ) = 5.93 \, \text{m/s}^2. 2. Apply kinematic equation Use v_f^2 = v_i^2 + 2a \cdot d with v_i = 0, a = 5.93 \, \text{m/s}^2, and d = 30.0 \, \text{m}. Therefore, v_f^2 = 0 + 2 \cdot 5.93 \cdot 30.0. 3. Solve for final velocity v_f = \sqrt{2 \cdot 5.93 \cdot 30.0}.

Explanation

1. Calculate the acceleration<br /> Use $a = g \cdot \sin(\theta)$ where $g = 9.81 \, \text{m/s}^2$ and $\theta = 37.2^\circ$. Thus, $a = 9.81 \cdot \sin(37.2^\circ) = 5.93 \, \text{m/s}^2$.<br />2. Apply kinematic equation<br /> Use $v_f^2 = v_i^2 + 2a \cdot d$ with $v_i = 0$, $a = 5.93 \, \text{m/s}^2$, and $d = 30.0 \, \text{m}$. Therefore, $v_f^2 = 0 + 2 \cdot 5.93 \cdot 30.0$.<br />3. Solve for final velocity<br /> $v_f = \sqrt{2 \cdot 5.93 \cdot 30.0}$.
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