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Chapter 4: Problem 4

Starting from rest, a car accelerates at \(2.0 \mathrm{m} / \mathrm{s}^{2}\) up a hill that is inclined \(5.5^{\circ}\) above the horizontal. How far (a) horizontally and (b) vertically has the car traveled in 12 s?

Short Answer

Expert verified
Horizontal distance is approximately 143.3 m and vertical distance is approximately 13.8 m.

Step by step solution

01

Recognize the Given Variables

The problem describes a car starting from rest. Its acceleration is given as \( a = 2.0 \, \text{m/s}^2 \), time \( t = 12 \, \text{s} \), and the inclination of the hill is \( \theta = 5.5^\circ \). We need to find the horizontal and vertical distances after 12 seconds.
02

Use Kinematic Equation for Total Distance

To find the total distance traveled along the inclined hill, we use the kinematic equation: \( s = ut + \frac{1}{2}at^2 \). Since the car starts from rest, \( u = 0 \). So, \( s = \frac{1}{2} \times 2.0 \, \text{m/s}^2 \times (12 \, \text{s})^2 \). Calculate \( s \).
03

Calculate Total Distance

Compute \( s = \frac{1}{2} \times 2.0 \times 144 = 144 \, \text{m} \). This is the distance the car traveled along the incline.
04

Find Horizontal Distance

The horizontal distance \( d_x \) can be found using the cosine component. Use the formula \( d_x = s \cdot \cos(\theta) \). Calculate \( d_x = 144 \, \text{m} \times \cos(5.5^\circ) \).
05

Calculate Horizontal Distance

Using a calculator, find \( \cos(5.5^\circ) \approx 0.995 \), so \( d_x = 144 \times 0.995 \approx 143.3 \, \text{m} \).
06

Find Vertical Distance

The vertical distance \( d_y \) is calculated using the sine component: \( d_y = s \cdot \sin(\theta) \). Calculate \( d_y = 144 \, \text{m} \times \sin(5.5^\circ) \).
07

Calculate Vertical Distance

Find \( \sin(5.5^\circ) \approx 0.096 \), so \( d_y = 144 \times 0.096 \approx 13.8 \, \text{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
In kinematics, acceleration represents how quickly the velocity of an object changes over time. It is a vector, meaning it has both magnitude and direction. In the problem, the car accelerates at a constant rate of \(2.0 \, \text{m/s}^2\). This constant acceleration indicates that the velocity of the car increases steadily every second. To calculate the effect of this acceleration over time, we use the kinematic equation which includes acceleration, initial velocity, and time. Since the car starts from rest, its initial velocity is zero, simplifying calculations.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, helping us understand how gravity and forces interact with motion. In the context of this problem, the car tackles a slope of \(5.5^\circ\) relative to the horizontal plane. The incline affects how we calculate the distances traveled along the horizontal and vertical axes, leveraging trigonometric functions to account for the angle. The car's path, in this case, is affected by gravitational forces breaking down into components parallel and perpendicular to the incline.
Kinematic Equations
Kinematic equations are essential tools for solving motion-related problems without regard to the forces creating that motion. They allow us to relate quantities like displacement, initial velocity, final velocity, acceleration, and time. In the exercise, we apply the equation \(s = ut + \frac{1}{2}at^2\), where \(u = 0\) since the car starts from rest. The equation helps determine the total distance the car travels along the incline in a given timeframe. The distance calculated is 144 meters, which serves as the basis for finding the horizontal and vertical distances using trigonometry.
Trigonometry in Physics
Trigonometry is often crucial in physics, especially when analyzing movements on an angle. In this exercise, we used trigonometric functions (cosine and sine) to break down the total distance (144 meters) into its horizontal and vertical components. Using \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\), the horizontal distance is found by multiplying the total distance with \(\cos(5.5^\circ)\). Similarly, \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\) gives us the vertical distance when multiplied by the total distance. These calculations reveal how trigonometry allows complex motions to be analyzed in simpler components.

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Most popular questions from this chapter

Suppose we change the dolphin's launch angle to \(45.0^{\circ}\), but everything else remains the same. Thus, the horizontal distance to the ball is \(5.50 \mathrm{m}\), the drop height is \(4.10 \mathrm{m},\) and the dolphin's launch speed is \(12.0 \mathrm{m} / \mathrm{s}\). (a) What is the vertical distance between the dolphin and the ball when the dolphin reaches the horizontal position of the ball? We refer to this as the "miss distance." (b) If the dolphin's launch speed is reduced, will the miss distance increase, decrease, or stay the same? (c) Find the miss distance for a launch speed of \(10.0 \mathrm{m} / \mathrm{s}\).

As you walk briskly down the street, you toss a small ball into the air. (a) If you want the ball to land in your hand when it comes back down, should you toss the ball straight upward, in a forward direction, or in a backward direction, relative to your body? (b) Choose the best explanation from among the following: I. If the ball is thrown straight up you will leave it behind. II. You have to throw the ball in the direction you are walking. III. The ball moves in the forward direction with your walking speed at all times.

On a hot summer day, a young girl swings on a rope above the local swimming hole (Figure \(4-20\) ). When she lets go of the rope her initial velocity is \(2.25 \mathrm{m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) above the horizontal. If she is in flight for \(0.616 \mathrm{s}\), how high above the water was she when she let go of the rope?

A golfer gives a ball a maximum initial speed of \(34.4 \mathrm{m} / \mathrm{s}\). (a) What is the longest possible hole-in-one for this golfer? Neglect any distance the ball might roll on the green and assume that the tee and the green are at the same level. (b) What is the minimum speed of the ball during this hole-in-one shot?

Two divers run horizontally off the edge of a low cliff. Diver 2 runs with twice the speed of diver 1 . (a) When the divers hit the water, is the horizontal distance covered by diver 2 twice as much, four times as much, or equal to the horizontal distance covered by diver \(1 ?\) (b) Choose the best explanation from among the following: I. The drop time is the same for both divers. II. Drop distance depends on \(t^{2}\). III. All divers in free fall cover the same distance.
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