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A car starts from rest and travels for \(5.0 \mathrm{~s}\) with a uniform acceleration of \(+1.5 \mathrm{~m} / \mathrm{s}^{2} .\) The driver then applies the brakes, causing a uniform acceleration of \(-2.0 \mathrm{~m} / \mathrm{s}^{2}\). If the brakes are applied for \(3.0 \mathrm{~s}\), (a) how fast is the car going at the end of the braking period, and (b) how far has the car gone?

Short Answer

Expert verified
The final velocity of the car is \(1.5 m/s\) and the total distance covered by the car is \(33.75m\).

Step by step solution

01

Identify Variables for First Phase of Motion

Firstly, identify the given variables (initial velocity, time, acceleration) for the initial acceleration phase. The car starts from rest so the initial velocity (\(u\)) is 0 m/s. It travels for a time (\(t\)) of \(5.0s\) with a uniform acceleration (\(a\)) of \(+1.5 m/s^{2}\).
02

Calculate Final Velocity after Initial Acceleration

Use the formula for final velocity (\(v\)) given initial velocity, acceleration, and time: \( v = u + at \). Substitute the known values from step 1 to get: \( v = 0m/s + 1.5 m/s^{2} x 5.0s = 7.5 m/s\). So, the car reaches a velocity of \(7.5 m/s\) after \(5s\).
03

Identify Variables for Second Phase of Motion

Now, consider the given variables for the deceleration phase: the brakes are applied causing a negative acceleration of \(-2.0 m/s^{2}\) and are applied for \(3.0s\). The initial velocity for this phase would be the final velocity from the earlier phase i.e. \(7.5 m/s\).
04

Calculate Final Velocity after Brakes are Applied

Use the same formula for final velocity again, this time with the values from the braking phase. Substitute into the equation: \(v = u + at\) to get \(v = 7.5m/s -2.0 m/s^{2} x 3.0s = 1.5 m/s\). So, the car is going at a velocity of \(1.5 m/s\) at the end of the braking period.
05

Calculate Overall Distance Covered

To find the total distance covered, sums of distances covered in both the acceleration and deceleration phase need to be calculated. Use the formula for distance covered given the initial velocity, acceleration, and time: \(s = ut + 0.5at^{2}\). The distance covered in the first phase is: \(s = 0.0m/s x 5.0s + 0.5(1.5 m/s^{2})(5.0s)^{2} = 18.75m \), and in the second phase: \(s = 7.5m/s x 3.0s + 0.5 (-2.0 m/s^{2}) (3.0s)^{2} = 15.0m\). Adding these two distances gives the total distance the car has travelled: \(18.75m + 15m = 33.75m\).

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Key Concepts

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

Final Velocity Calculation
Understanding the final velocity of a moving object is fundamental when studying the motions under uniform acceleration. It's the velocity which the object will have after a certain period of applying a constant acceleration. Imagine trying to gauge just how fast a car is going after steadily speeding up or slowing down. It's this concept that helps us analyze such scenarios.

To calculate the final velocity, we use the equation:
\[v = u + at\]
Here,
  • \(v\) represents the final velocity,
  • \(u\) is the initial velocity,
  • \(a\) stands for acceleration, and
  • \(t\) is the time over which this acceleration is applied.

Once we've established all the necessary values, it's a straightforward substitution to find the car's velocity at any given time. As demonstrated in the problem's step-by-step solution, by knowing the car's initial speed, the time duration, and the consistent acceleration, we can find out just how fast the car is travelling after the application of the acceleration. This becomes particularly useful not only in academic problems but also in real-life situations such as calculating stopping distances for vehicles or determining the speeds in collision reconstructions.
Distance Travelled Calculation
To grasp the trajectory of an object or vehicle over a duration, it's essential to calculate the distance it has travelled. This component of kinematic analysis allows us to predict positions or plan routes effectively.

The foundational formula to calculate the distance (\(s\)) covered under uniform acceleration is:
\[s = ut + \frac{1}{2}at^2\]
This formula is a pivotal part of physics since it amalgamates initial velocity (\(u\)), acceleration (\(a\)), and time (\(t\)) to reveal the distance travelled. For our scenario, remember:
  • \(s\) is the distance travelled,
  • \(u\) is the initial velocity,
  • \(a\) is the acceleration, and
  • \(t\) is the time.

Carefully substituting the known quantities into this equation allows for a step-by-step breakdown of the distance covered during each phase of motion—first, as the car accelerates away from a resting position, and then as it decelerates upon braking. Summing these distances from both the accelerating and braking phases, as illustrated in the provided solution, yields the total distance the car has journeyed.
Kinematics in One Dimension
The branch of physics called kinematics deals with the motion of objects without considering the forces that cause the motion. Within kinematics, the simple case of one-dimensional motion is often where the fundamentals are applied. It encompasses objects moving in a straight line with uniform acceleration, such as a car on a straight track.

In one-dimensional kinematics, we rely on a few central equations to describe motion completely. These include equations for velocity, as well as distance travelled, as introduced in the previous sections. A clear understanding of kinematics in one dimension allows you to predict future motion, analyze current motion, and reconstruct past motion from given information.

When focusing on one-dimensional motion, you'll often use:
  • The formula for final velocity to gauge speed at a given moment,
  • The distance formula to calculate how far the object has traveled over time, and
  • Equations for time and acceleration, if they are unknown variables.

This streamlined approach simplifies complex motions into manageable calculations and fosters a deeper understanding of motion overall. Whether solving textbook problems or deciphering real-world dynamics, the principles of one-dimensional kinematics are versatile tools in the physicist's arsenal.

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Most popular questions from this chapter

A tennis player tosses a tennis ball straight up and then catches it after \(2.00 \mathrm{~s}\) at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? (b) What is the velocity of the ball when it reaches its maximum height? Find (c) the initial velocity of the ball and (d) the maximum height it reaches.

In 1865 Jules Verne proposed sending men to the Moon by firing a space capsule from a 220 -m-long cannon with final speed of \(10.97 \mathrm{~km} / \mathrm{s}\). What would have been the unrealistically large acceleration experienced by the space travelers during their launch? (A human can stand an acceleration of \(15 g\) for a short time.) Compare your answer with the free- fall acceleration, \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

Speedy Sue, driving at \(30.0 \mathrm{~m} / \mathrm{s}\), enters a one-lane tunnel. She then observes a slow-moving van \(155 \mathrm{~m}\) ahead traveling at \(5.00 \mathrm{~m} / \mathrm{s}\). Sue applies her brakes but can accelerate only at \(-2.00 \mathrm{~m} / \mathrm{s}^{2}\) because the road is wet. Will there be a collision? State how you decide. If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach betwecn Sue's car and the van.

A train \(400 \mathrm{~m}\) long is moving on a straight track with a speed of \(82.4 \mathrm{~km} / \mathrm{h}\). The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of \(16.4 \mathrm{~km} / \mathrm{h}\). Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.

A ball is thrown upward from the ground with an initial speed of \(25 \mathrm{~m} / \mathrm{s}\); at the same instant, another ball is dropped from a building \(15 \mathrm{~m}\) high. After how long will the balls be at the same height?

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