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A truck on a straight road starts from rest, accelerating at \(2.00 \mathrm{m} / \mathrm{s}^{2}\) until it reaches a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) Then the truck travels for \(20.0 \mathrm{s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.00 \mathrm{s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

Short Answer

Expert verified
The truck is in motion for a total time of 35.0 s. The average velocity of the truck over the course of the motion is 15.7 m/s.

Step by step solution

01

Calculate the time taken to reach maximum speed

The time for the truck to reach a speed of \(20.0 m/s\) with a constant acceleration of \(2.00 m/s^2\) is given by the formula \(t = v/a\). Substitute \(v = 20.0 m/s\) and \(a = 2.00 m/s^2\) into the formula, resulting in \(t = 20.0 m/s ÷ 2.00 m/s^2 = 10.0 s\).
02

Total time the truck in motion

The total time for the truck in motion is the sum of the time taken to reach maximum speed, the time cruising at constant speed, and the time to stop. This results in a total time of \(10.0 s + 20.0 s + 5.0 s = 35.0 s\).
03

Calculate the distance during acceleration

The distance travelled while the truck is accelerating is given by the equation \(d = 0.5at^2\). Substitute \(a = 2.00 m/s^2\) and \(t = 10.0 s\) into the formula, getting \(d = 0.5 * 2.00 m/s^2 * (10.0 s)^2 = 100.0 m\).
04

Calculate the distance during cruising

The distance travelled while the truck is cruising is given by \(d = vt\). Substituting \(v = 20.0 m/s\) and \(t = 20.0 s\) into the formula, we get \(d = 20.0 m/s * 20.0 s = 400.0 m\).
05

Calculate the distance during deceleration

The distance travelled while the truck is decelerating can be found using the formula \(d = vt - 0.5at^2\). Here, \(v = 20.0 m/s\), \(a = 20.0 m/s^2 / 5.0 s = 4.00 m/s^2\) (using \(v = at\), a variant of the equation used in Step 1), and \(t = 5.0 s\). Substituting these into the formula gives \(d = (20.0 m/s * 5.0 s) - 0.5 * 4.00 m/s^2 * (5.0 s)^2 = 50.0 m\).
06

Calculate the average velocity

The average velocity is given by the total displacement divided by the total time. The total displacement is the sum of the distances calculated in Steps 3, 4 and 5, and gives \(100.0 m + 400.0 m + 50.0 m = 550.0 m\). The total time was calculated in Step 2 as 35.0 s. So, the average velocity is \(550.0 m ÷ 35.0 s ≈ 15.7 m/s\).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a constant rate. This means that for every equal time interval, the change in velocity is the same. In the exercise, the truck accelerates uniformly from rest, meaning it starts at zero velocity. By accelerating at a constant rate of 2.00 m/s², the truck increases its velocity by 2 meters per second for every second it is in motion.
This situation is typical of a straight-line motion where external conditions, like road conditions and engine power, are constant. A familiar equation used to describe this process is \( v = at \), where \( v \) is the final velocity, \( a \) is the acceleration, and \( t \) is the time. It shows how velocity builds over time when acceleration doesn't change.
  • Uniform acceleration ensures predictable motion patterns.
  • This scenario can only occur when there is no variation in the force applied to the object or when friction is constant.
Understanding uniform acceleration is critical as it leads to easier calculations of various motion parameters using kinematic equations.
Average Velocity
Average velocity is a measure of the total displacement over the total time taken. In simpler terms, it tells us the overall speed and direction of an object during its motion. Unlike instantaneous velocity, which looks at a particular moment, average velocity considers the motion from start to finish.
In our truck exercise, the average velocity was needed to summarize the movement throughout the entire trip, which included acceleration, constant speed, and deceleration phases.
The formula to calculate average velocity is \( \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} \). For the truck, which displaced 550 meters over 35 seconds, the average velocity came out to 15.7 m/s.
  • Average velocity gives a broad overview of motion.
  • It is significant in finding efficiency and travel analysis.
Applying the concept of average velocity allows for understanding how effective the movement was, even if the truck varied its speed throughout the trip.
Motion Equations
Motion equations, also known as kinematic equations, describe the mathematical relationships between different aspects of motion under uniform acceleration. These equations are instrumental in solving various real-world physics problems as they tie together time, velocity, displacement, and acceleration.
In our exercise, a few key motion equations were used, including those for calculating time under acceleration \( t = \frac{v}{a} \), and displacement \( d = vt \) during cruising, and \( d = 0.5at^2 \) while accelerating or decelerating.
The flexibility and utility of these equations make them essential for accurately describing motion, whether it's a moving vehicle or a freely falling object.
  • Motion equations assist in predicting future states of motion.
  • They are foundational in analyses from engineering to daily applications like transport and sports.
Being comfortable with motion equations allows students to approach complex motion scenarios with confidence.

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Most popular questions from this chapter

A particle moves along the \(x\) axis according to the equation \(x=2.00+3.00 t-1.00 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. At \(t=3.00 \mathrm{s},\) find (a) the position of the particle, (b) its velocity, and (c) its acceleration.

A speedboat moving at \(30.0 \mathrm{m} / \mathrm{s}\) approaches a no-wake buoy marker \(100 \mathrm{m}\) ahead. The pilot slows the boat with a constant acceleration of \(-3.50 \mathrm{m} / \mathrm{s}^{2}\) by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?

A commuter train travels between two downtown stations. Because the stations are only \(1.00 \mathrm{km}\) apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the time interval \(\Delta t\) between two stations by accelerating for a time interval \(\Delta t_{1}\) at a rate \(a_{1}=0.100 \mathrm{m} / \mathrm{s}^{2}\) and then immediately braking with acceleration \(a_{2}=-0.500 \mathrm{m} / \mathrm{s}^{2}\) for a time interval \(\Delta t_{2} .\) Find the minimum time interval of travel \(\Delta t\) and the time interval \(\Delta t_{1}\).

An inquisitive physics student and mountain climber climbs a 50.0 -m cliff that overhangs a calm pool of water. He throws two stones vertically downward, \(1.00 \mathrm{s}\) apart, and observes that they cause a single splash. The first stone has an initial speed of \(2.00 \mathrm{m} / \mathrm{s} .\) (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water?

A particle moves according to the equation \(x=10 t^{2}\) where \(x\) is in meters and \(t\) is in seconds. (a) Find the average velocity for the time interval from \(2.00 \mathrm{s}\) to \(3.00 \mathrm{s}\). (b) Find the average velocity for the time interval from 2.00 to \(2.10 \mathrm{s}\).

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