/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A car is parked on a steep incli... [FREE SOLUTION] | 91Ó°ÊÓ

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A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of \(37.0^{\circ}\) below the horizontal. The negligent driver leaves the car in neutral, and the parking brakes are defective. Starting from rest at \(t=0,\) the car rolls down the incline with a constant acceleration of \(4.00 \mathrm{m} / \mathrm{s}^{2},\) traveling \(50.0 \mathrm{m}\) to the edge of a vertical cliff. The cliff is \(30.0 \mathrm{m}\) above the ocean. Find (a) the speed of the car when it reaches the edge of the cliff and the time at which it arrives there, (b) the velocity of the car when it lands in the ocean, (c) the total time interval that the car is in motion, and (d) the position of the car when it lands in the ocean, relative to the base of the cliff.

Short Answer

Expert verified
a) The speed of the car when it gets to the edge of the cliff is \(v_{f_{incline}} = 20.0 \, m/s\), and time is \(t_{incline} = 5.0 \, s\). b) The velocity of the car when it hits the water is \(v_{f_{fall}} = 24.5 \, m/s\) downward. c) The total time interval the car is in motion is \(t_{total} = 7.9 \, s\). d) The position of the car when it hits the water, relative to the base of the cliff, is 41.0 meters away from the cliff.

Step by step solution

01

Calculate the speed of the car when it reaches the edge of the cliff

The final velocity of the car can be computed using the equation of motion: \(v_f = v_i + at\), where \(v_f\) denotes the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. However, time is unknown. Another equation is needed to eliminate \(t\): \(d = v_i t + \frac{1}{2} a t^2\), where \(d\) denotes distance. From here, solve for \(t\) and substitute in the first equation: \(t = \sqrt{\frac{2d}{a}}\). Substituting \(t\) into the first equation, the final velocity is found to be \(v_f = \sqrt{2ad}\). By substituting \(a = 4.00 \, m/s^{2}\) and \(d = 50.0 \, m\), the final velocity when the car reaches the cliff edge can be obtained.
02

Determine the time the car arrives at the edge

From the equation in step 1, \(t = \sqrt{\frac{2d}{a}}\), substituting \(d = 50.0 \, m\) and \(a = 4.00 \, m/s^{2}\), you can get the time it takes for the car to get the edge of the cliff.
03

Calculate the final velocity of the car as it hits the ocean

The final velocity can be found by using the equation: \(v_f = v_i + gt\), where \(g\) denotes the acceleration due to gravity (\(9.8 \, m/s^{2}\)) and \(t\) denotes the time it takes for the car to fall from the cliff to the ocean. However, \(t\) is unknown at this point. Another equation is required to eliminate \(t\): We can use \(h = v_i t + \frac{1}{2}gt^2\), where \(h\) denotes the height of the cliff (\(30.0 \, m\)). Here, \(v_i\) is the initial vertical velocity when the car leaves the cliff's edge, and it's zero because the cliff's edge is horizontal. Solving this for \(t\), \(t = \sqrt{\frac{2h}{g}}\). Substituting \(t\) into the first equation, and simplifying, you can find the vertical component of the final velocity.
04

Calculate the total time car is in motion

The total time the car is in motion is the sum of the time it takes to roll down the incline and the time it falls from the cliff to the ocean. Add the time calculated in Step 2 and the time calculated in Step 3.
05

Find the position of the car when it lands in the ocean

The horizontal displacement of the car when it hits the water can be determined by using the equation of motion: \(d = v_i t + \frac{1}{2} a t^2\). The initial vertical velocity of the car when leaving the cliff is zero, thus the displacement is determined by \(d = vt\), where \(v\) is the horizontal component of the velocity and \(t\) is the time it takes for the car to fall from the cliff to the ocean.

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Key Concepts

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

Projectile Motion
Projectile motion is a fascinating concept often found in physics. It's simply the motion of an object that is thrown or projected into the air and is subject to the acceleration due to gravity. When we tackle problems involving projectile motion, like the car rolling off the cliff in the exercise, we usually separate the motion into horizontal and vertical components.
This separation is crucial because each component behaves independently. For the horizontal component, the acceleration is usually zero, which means the velocity remains constant. In our car exercise, once the car moves off the cliff edge, its horizontal speed doesn’t change until it hits the ocean.
For the vertical motion component, gravity is the one in control. It causes the object to speed up as it descends. Think of it this way: when the car rolls off the edge of the cliff, it starts descending towards the ocean, accelerating under gravity at 9.8 m/s². By analyzing each component separately, you can solve for various parameters, like the time of flight or impact velocity.
Inclined Plane
An inclined plane is simply a flat surface tilted at an angle, different from horizontal. It’s crucial to understand how an object behaves on an inclined plane because the gravity doesn’t work entirely as it does on flat ground. Let’s talk about what this means using our example of the car on an incline.
When the car rolls down, gravity tries to pull it straight down, but because of the incline, the car doesn't fall directly downwards. Instead, it accelerates along the slope. Two components arise from gravity: one acting parallel to the surface of the incline, and another acting perpendicular to it. The parallel component causes the car to speed up down the slope, while the perpendicular component keeps it pressed against the incline without affecting its speed.
  • The strength of the car's acceleration down the incline is determined by this parallel component (\(a = g \sin \theta\), where \(\theta\) is the incline angle).
  • For the parked car, the incline contributes to a constant acceleration value of 4.00 m/s² as the car rolls down the hill.
Understanding inclined planes helps break down these forces and solve for the car’s speed and the time it takes to reach the cliff's edge.
Constant Acceleration
Constant acceleration is a key concept in kinematics, where an object's velocity changes at a steady rate over time. In the car's situation on the incline, the acceleration doesn’t change as it rolls down, because we’re dealing with a frictionless surface and no brakes.
The equations of motion are our tools for solving problems with constant acceleration. These equations relate several aspects of the object’s motion: initial velocity, final velocity, acceleration, time, and displacement.
  • The basic formula is \(v_f = v_i + at\), which gives us the final velocity after a time \(t\) with acceleration \(a\).
  • Another crucial formula is \(d = v_i t + \frac{1}{2} a t^2\). This relates distance traveled to the other variables.
In the exercise, using \(a = 4.00\, m/s^2\) and \(d = 50.0\, m\), these equations help us find how fast the car is moving when it reaches the cliff, and the time required to get there. By mastering constant acceleration, you can tackle a variety of kinematic problems with confidence.

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Most popular questions from this chapter

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