/*! 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 1 A car starts at the origin and m... [FREE SOLUTION] | 91Ó°ÊÓ

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A car starts at the origin and moves with velocity \(\vec{v}=\) \((10 \mathrm{m} / \mathrm{s},\) northeast). How far from the origin will the car be after traveling for 45 s?

Short Answer

Expert verified
The car will be 450 meters away from the origin after 45 seconds of travel.

Step by step solution

01

- Understanding the given values

We are given that the velocity of the car is 10 m/s acting towards northeast and time is 45 seconds. We are asked to find how far the car will be from origin.
02

- Applying the formula

The distance traveled by an object with constant velocity is given by the product of the velocity and time. Hence, the formula for the distance is \(d = v \times t \) where \(d\) is the distance, \(v\) is the velocity and \(t\) is the time.
03

- Calculating the distance

Now, substituting the given values into the formula, we get \(d = 10\,m/s \times 45\,s = 450\,m\).

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Key Concepts

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

Velocity
Velocity is a key concept in kinematics. It describes how fast something is moving and in which direction. Think of it like a car racing down a track: speed tells you how fast the car is moving, but velocity tells you both its speed and its direction.

Velocity is represented as a vector, meaning it has both magnitude (how much) and direction. In the exercise, our car has a velocity of 10 m/s going northeast. This tells us not only the speed but also the exact path the car is following.

Understanding velocity is crucial for predicting future positions of a moving object. Whether calculating the car's future location or planning a safe route, knowing both the speed and direction helps make better decisions.
Distance Calculation
To find out how far an object travels, we use the formula: \[d = v \times t\] where \(d\) is the distance, \(v\) is velocity, and \(t\) is time. This simple multiplication gives us the total distance covered.

In our exercise, the car’s velocity is given as 10 m/s, and the time is 45 seconds. By multiplying these two values, \(d =10 \times 45\), we find that the car travels 450 meters from the origin.

Remember: when discussing motion along a straight path with constant velocity, this formula is your go-to tool. It works for any situation where you have a constant speed and a known time duration. Just plug in your values and compute the distance!
Constant Velocity Motion
Constant velocity means that an object is moving at an unchanged speed and direction. There are no sudden starts, stops, or turns, which simplifies calculations significantly.

In exercises like this one, constant velocity makes it easy to predict future positions. Using a consistent speed ensures the relationship between velocity, distance, and time remains constant.
  • No acceleration or deceleration to consider.
  • Simple arithmetic gets the job done.
  • Predictable and straightforward motion.
Once you understand that the car is moving at a constant velocity, you know that every second it covers exactly the same amount of ground. This is why the distance calculation is straightforward: just multiply the velocity by the time to find how far the car goes.

Understanding constant velocity makes every calculation more intuitive, reducing the need for complex equations and allowing us to focus on mastering the basics of kinematic movement.

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Most popular questions from this chapter

A \(1000 \mathrm{kg}\) weather rocket is launched straight up. The rocket motor provides a constant acceleration for \(16 \mathrm{s}\), then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. a. What was the rocket's acceleration during the first 16 s? b. What is the rocket's speed as it passes through a cloud \(5100 \mathrm{m}\) above the ground?

A rubber ball is shot straight up from the ground with speed \(v_{0}\) Simultaneously, a second rubber ball at height \(h\) directly above the first ball is dropped from rest. a. At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of \(v_{0}\) and \(g\). b. What is the maximum value of \(h\) for which a collision occurs before the first ball falls back to the ground? c. For what value of \(h\) does the collision occur at the instant when the first ball is at its highest point?

a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? b. What fraction of \(g\) is this? c. How far has the car traveled when it reaches 60 mph? Give your answer both in SI units and in feet.

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at \(30 \mathrm{m} / \mathrm{s}^{2}\) for \(30 \mathrm{s}\), then runs out of fuel. Ignore any air resistance effects. a. What is the rocket's maximum altitude? b. How long is the rocket in the air before hitting the ground? c. Draw a velocity-versus-time graph for the rocket from liftoff until it hits the ground.

An object starts from rest at \(x=0\) m at time \(t=0\) s. Five seconds later, at \(t=5.0 \mathrm{s},\) the object is observed to be at \(x=40.0 \mathrm{m}\) and to have velocity \(v_{x}=11 \mathrm{m} / \mathrm{s}\) a. Was the object's acceleration uniform or nonuniform? Explain your reasoning. b. Sketch the velocity-versus-time graph implied by these data. Is the graph a straight line or curved? If curved, is it concave upward or downward?

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