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Chapter 2: Problem 37

A rock is thrown straight up with an initial speed of \(24.0 \mathrm{~m} / \mathrm{s}\) Neglect air resistance. (a) At \(t=1.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing? (b) At \(t=3.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing?

Short Answer

Expert verified
(a) At \( t=1.0s \), the direction of velocity is upwards and the acceleration is downwards. The speed of the rock is decreasing. (b) At \( t=3.0s\), the direction of both velocity and acceleration is downwards. The speed of the rock is increasing.

Step by step solution

01

Determine the initial conditions

The rock is thrown straight up with an initial speed of 24.0 m/s. We are aware that the acceleration due to gravity, typically denoted as \(g\), is a constant -9.8 m/s虏. The negative sign signifies that it acts in the downward direction, contrary to the initial direction of throw.
02

Compute the velocity at \( t=1.0s \) and \( t=3.0s \)

The velocity \(v\) at any time \(t\) can be calculated using the formula \(v = u - g \cdot t\), where \(u\) is the initial speed and \(g\) is the acceleration due to gravity. So, at \(t = 1.0s\), \(v_1 = 24.0m/s - 9.8 m/s^2 \cdot 1.0s\) and at \(t = 3.0s\), \(v_2 = 24.0m/s - 9.8 m/s^2 \cdot 3.0s \).
03

Determine the direction and rate of change of speed

At \( t=1.0s \), as \(v_1\) is positive, the direction of the velocity is upwards and the rock is still moving upward. However, as it's less than the initial speed, the speed of the rock is decreasing. At \( t=3.0s \), as \(v_2\) is negative, the direction of the velocity is now downwards indicating that the rock is moving downwards. As the absolute speed (disregarding direction) is greater than at \(t=1.0s\), the speed of the rock is increasing again.
04

Determine acceleration direction

The acceleration due to gravity is always directed downwards (negative). Thus, even when the rock is thrown upwards and is temporarily moving against gravity, the acceleration is still acting downwards.

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Key Concepts

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

Initial Velocity
Initial velocity is a critical parameter in the study of motion, representing the speed and direction of an object at the start of analysis. For example, if you toss a rock upwards, the initial velocity is the speed at which it leaves your hand. It's important to note the direction as well - in this case, the initial velocity is upwards.

Imagine you're a quarterback throwing a football - the speed and angle at which you throw are the initial conditions setting the stage for the football's trajectory. In physics problems, initial velocity (\(u\text{ or }v_0\text{, depending on notation}\)) can influence time of flight, maximum height, and final velocity based on equations of motion.

Acceleration Due to Gravity
Acceleration due to gravity, denoted as \(g\), is the acceleration that the Earth imparts to objects falling towards it without other force acting, like air resistance. It's a constant value of approximately \(9.8 \text{m/s}^2\) on the Earth's surface, always acting downwards, towards the center of the Earth.

To conceptualize this, imagine dropping an apple; it speeds up as it falls because of gravity. Regardless of the initial motion of the object, whether thrown upwards, downwards, or horizontally, gravity consistently acts to accelerate it downwards. This characteristic is a cornerstone of kinematics as it allows us to predict the motion of projectiles and freely falling objects.

Velocity-Time Relationship
The velocity-time relationship is one of the foundational aspects of kinematics, capturing how an object's speed changes over time. In our context, the velocity of a thrown rock is governed by the equation \(v = u - g \cdot t\), where \(v\) represents the final velocity at any given time \(t\), \(u\) is the initial velocity, and \(g\) is the acceleration due to gravity. This equation reflects the effect of gravity on the rock over time.

As time progresses, gravity reduces the rock's upward velocity up until it reaches its peak, where the velocity is zero. Then, gravity accelerates the rock downwards, increasing its velocity in the downward direction. The direction of velocity changes once the rock reaches its highest point, a crucial insight for understanding the motion of objects under the influence of gravity.

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Most popular questions from this chapter

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than \(1.00 \mathrm{~s}\) in the air (their "hang time"). Treat the athlete as a particle and let \(y_{\max }\) be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above \(y_{\max } / 2\) to the time it takes him to go from the floor to that height. Ignore air resistance.

The acceleration of a motorcycle is given by \(a_{x}(t)=A t-B t^{2},\) where \(A=1.50 \mathrm{~m} / \mathrm{s}^{3}\) and \(B=0.120 \mathrm{~m} / \mathrm{s}^{4} .\) The motorcycle is at rest at the origin at time \(t=0\). (a) Find its position and velocity as functions of time. (b) Calculate the maximum velocity it attains.

A turtle crawls along a straight line, which we'll call the \(x\) -axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{~cm}+\) \((2.00 \mathrm{~cm} / \mathrm{s}) t-\left(0.0625 \mathrm{~cm} / \mathrm{s}^{2}\right) t^{2} .\) (a) Find the turtle's initial velocity. initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of \(10.0 \mathrm{~cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of \(x\) versus \(t, v_{1}\) versus \(t,\) and \(a_{f}\) versus \(t,\) for the time interval \(t=0\) to \(t=40 \mathrm{~s}\)

The fastest measured pitched baseball left the pitcher's hand at a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). If the pitcher was in contact with the ball over a distance of \(1.50 \mathrm{~m}\) and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?
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