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An observer by the side of a straight, level, northsouth road watches a car (A) moving south at a rate of \(75 \mathrm{~km} / \mathrm{h}\). A driver in another car (B) going north at \(50 \mathrm{~km} / \mathrm{h}\) also observes car \(\mathrm{A}\). (a) What is car A's velocity as observed from car B? (Take north to be positive.) (b) If the roadside observer sees car A brake to a stop in \(6.0 \mathrm{~s}\) what constant acceleration would be measured? (c) What constant acceleration would the driver in car B measure for the braking car A?

Short Answer

Expert verified
(a) -125 km/h, (b) 3.47 m/s², (c) 3.47 m/s².

Step by step solution

01

Understanding Relative Velocity

To find the velocity of car A as observed from car B, we need to calculate the relative velocity of car A with respect to car B. This is done using the formula: \( v_{AB} = v_A - v_B \). Both velocities are given with respect to a stationary observer on the roadside.
02

Calculate Individual Velocities

The velocity of car A \( v_A \) is \(-75 \text{ km/h}\) because it moves south. For car B, the velocity \( v_B \) is \(50 \text{ km/h}\) because it moves north. The sign indicates direction with north positive and south negative.
03

Calculate Relative Velocity of A with Respect to B

Substitute the velocities into the relative velocity formula: \( v_{AB} = (-75) - 50 = -125 \text{ km/h} \). This means car A appears to move at \(-125 \text{ km/h}\) as seen from car B.
04

Understanding Acceleration Calculation

Acceleration is calculated as the change in velocity over time. Since car A stops, its final velocity \( v_f = 0 \). Initial velocity \( v_i = -75 \text{ km/h} \). Convert to m/s: \(-75 \text{ km/h} = -20.83 \text{ m/s}\). Time \( t = 6.0 \text{ s} \).
05

Calculate Acceleration for Observer

Use the formula \( a = \frac{v_f - v_i}{t} \) to find the acceleration measured by the roadside observer: \( a = \frac{0 - (-20.83)}{6.0} \approx 3.47 \text{ m/s}^2 \).
06

Relative Acceleration from Car B’s Perspective

The acceleration experienced by car B for car A is the same as the roadside observer since acceleration is invariant under relative motion (for constant motion observers), thus remains \( a = 3.47 \text{ m/s}^2 \).

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Key Concepts

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

Constant Acceleration
Constant acceleration describes a steady change in velocity over time. When a vehicle moves with constant acceleration, the rate at which its speed changes remains the same. In the given problem, car A undergoes constant acceleration when it brakes to stop. This means car A's velocity decreases at a consistent rate until it reaches zero.

We calculate constant acceleration using the formula: \[ a = \frac{v_f - v_i}{t} \] where:
  • \(v_f\) is the final velocity
  • \(v_i\) is the initial velocity
  • \(t\) is the time period over which the change occurs
In car A's case, it goes from \(-20.83 \text{ m/s}\) to 0 in 6 seconds. Plugging these values into the formula gives an acceleration of approximately \(3.47 \text{ m/s}^2\).

The critical point to understand here is that even though car A's speed changes, the rate of this change—its acceleration—remains constant as it brakes.
Velocity Calculation
Velocity calculation involves determining the rate of change of position in a specific direction. It’s important to note both the magnitude and direction when discussing velocity, as it is a vector quantity. In the exercise, you’re asked to calculate the velocity of car A as observed by someone in car B.

To do this, we use the relative velocity formula, which helps find how fast one object is moving relative to another. The formula is: \[ v_{AB} = v_A - v_B \] Here,
  • \(v_{A}\) is the velocity of car A
  • \(v_{B}\) is the velocity of car B
Since both cars are moving in opposite directions, with north being positive, car A's velocity (southward) is \(-75 \text{ km/h}\) and car B's (northward) is \(50 \text{ km/h}\). When calculating \(v_{AB}\), it results in \(-125 \text{ km/h}\), meaning car A seems to move faster backward from car B's viewpoint.
Observer's Perspective
From an observer's perspective, understanding how motion appears depends greatly on their frame of reference. In the problem, there are two observers: one on the roadside and another in car B.

For the roadside observer, car A's deceleration is straightforward. They see it slowing down at a constant rate until it halts. The acceleration they observe is a direct calculation from how quickly car A's speed changes over the 6-second interval.

In contrast, the driver in car B experiences a more complex scenario. Not only do they perceive car A as moving in the opposite direction, but any changes in speed also consider car B's own velocity. However, when car A brakes, its acceleration magnitude remains the same from both perspectives because acceleration is constant irrespective of the observer’s motion, provided that both observers travel at uniform speeds. This invariance means that both observers measure the same acceleration value for car A.

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Most popular questions from this chapter

You are in a fast powerboat that is capable of a sustained steady speed of \(20.0 \mathrm{~m} / \mathrm{s}\) in still water. On a swift, straight section of a river you travel parallel to the bank of the river. You note that you take \(15.0 \mathrm{~s}\) to go between two trees on the riverbank that are \(400 \mathrm{~m}\) apart. (a) (1) Are you traveling with the current, (2) are you traveling against the current, or (3) is there no current? (b) If there is a current [reasoned in part (a)], determine its speed.

A golfer lines up for her first putt at a hole that is \(10.5 \mathrm{~m}\) exactly northwest of her ball's location. She hits the ball \(10.5 \mathrm{~m}\) and straight, but at the wrong angle, \(40^{\circ}\) from due north. In order for the golfer to have a "twoputt green," determine (a) the angle of the second putt and (b) the magnitude of the second putt's displacement. (c) Determine why you cannot determine the length of travel of the second putt.

A railroad flatbed car is set up for a physics demonstration. It is set to roll horizontally on its straight rails at a constant speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) On it is rigged a small launcher capable of launching a small lead ball vertically upward, relative to the bed of the car, at a speed of \(25.0 \mathrm{~m} / \mathrm{s} .\) (a) Compare [by making two sketches] the description of the motion from the point of view of two different observers: one riding on the car and one at rest on the ground next to the tracks. (b) How long does it take the ball to return to its launch location? (c) Compare the ball's velocity at the top of its motion from the viewpoint of each of the two observers and explain any differences. (d) What is the launch angle (relative to the ground) and speed of the ball according to the ground observer? (e) How far down the rails has the car moved when the ball lands back at the launcher? Compare this distance to how far the ball has moved relative to the car.

A ball rolling on a table has a velocity with rectangular components \(v_{x}=0.60 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=0.80 \mathrm{~m} / \mathrm{s} .\) What is the displacement of the ball in an interval of \(2.5 \mathrm{~s} ?\)

A shopper in a mall is on an escalator that is moving downward at an angle of \(41.8^{\circ}\) below the horizontal at a constant speed of \(0.75 \mathrm{~m} / \mathrm{s}\). At the same time a little boy drops a toy parachute from a floor above the escalator and it descends at a steady vertical speed of \(0.50 \mathrm{~m} / \mathrm{s}\). Determine the speed of the parachute toy as observed from the moving escalator.

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