Problem 19
Question
At a certain time a particle had a speed of \(18 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction, and \(2.4 \mathrm{~s}\) later its speed was \(30 \mathrm{~m} / \mathrm{s}\) in the opposite direction. What is the average acceleration of the particle during this \(2.4 \mathrm{~s}\) interval?
Step-by-Step Solution
Verified Answer
The average acceleration is
\(-20 \, m/s^2\).
1Step 1: Understand the Situation
The problem involves a particle that changes its velocity from a positive direction to a negative direction over a given time period. We need to calculate the average acceleration over this interval.
2Step 2: Identify Initial and Final Velocities
The initial velocity (\(v_i\)) of the particle is \(18 \, m/s\) in the positive x direction. The final velocity (\(v_f\)) is \(30 \, m/s\) in the opposite direction, which we'll consider as \(-30 \, m/s\).Thus, we have:\[v_i = 18 \, \text{m/s} , \, v_f = -30 \, \text{m/s}\]
3Step 3: Determine the Time Interval
The time interval over which this change occurs is \(2.4 \, s\). We denote this as \(t\).\[t = 2.4 \, \text{s}\]
4Step 4: Use the Average Acceleration Formula
The average acceleration (\(a_{avg}\)) can be calculated using the formula:\[a_{avg} = \frac{v_f - v_i}{t}\]Substitute the known values into the formula:\[a_{avg} = \frac{-30 \, \text{m/s} - 18 \, \text{m/s}}{2.4 \, \text{s}}\]
5Step 5: Calculate the Average Acceleration
Solve the equation from Step 4:\[a_{avg} = \frac{-30 - 18}{2.4}\]\[a_{avg} = \frac{-48}{2.4}\]\[a_{avg} = -20 \, \text{m/s}^2\]
6Step 6: Interpret the Result
The average acceleration of the particle is \(-20 \, m/s^2\), indicating that it speeds up in the negative direction, as expected since it ended up moving in the opposite direction to its initial motion.
Key Concepts
velocity changetime intervalkinematics
velocity change
Velocity change is a crucial part of understanding how objects move. When you look at an object's velocity, you see both the speed and direction it's traveling. A change in velocity means either a change in speed, a change in direction, or both. For our exercise, the particle starts at a speed of 18 m/s in the positive x direction and changes to 30 m/s in the negative x direction. Let's break this down:
- The initial velocity (\(v_i\)) is +18 m/s.
- The final velocity (\(v_f\)) is -30 m/s.
time interval
The time interval is the duration over which changes in motion occur. In physics, it's typically denoted by the letter \(t\). Knowing the time interval helps us understand how quickly or slowly a velocity change happens.In our problem, the time interval is given as 2.4 seconds. This is the duration from when the particle's initial velocity is measured to when the final velocity is taken. This crucial piece of information allows us to compute the average acceleration. When you calculate things like average acceleration, the time interval serves as the denominator in the equation: \(a_{avg} = \frac{\Delta v}{t}\).This means the longer the interval, assuming the same change in velocity, the smaller the average acceleration, and vice versa.
kinematics
Kinematics is a branch of physics dealing with motion without considering the forces that cause it. It focuses on parameters like velocity, acceleration, and time, helping us understand the movement patterns of objects.When solving our exercise, we are applying kinematic principles. We determined the average acceleration using the formula: \(a_{avg} = \frac{v_f - v_i}{t}\).This calculation hinges on understanding the change in velocity and the time over which this change occurs. In real-world applications, kinematics allows us to predict how an object will move under known conditions, aiding in fields like engineering, astronomy, and even sports science. So, by grasping the basic kinematic equations, we can predict and analyze motion with precision and clarity.
Other exercises in this chapter
Problem 15
(a) If a particlc's position is given by \(x=4-12 t+3 t^{2}\) (where \(t\) is in seconds and \(x\) is in meters), what is its velocity at \(t=1 \mathrm{~s} ?(\m
View solution Problem 18
The position of a particle moving along an \(x\) axis is given by \(x=12 t^{2}-2 t^{3}\), where \(x\) is in meters and \(t\) is in seconds. Determine (a) the po
View solution Problem 20
(a) If the position of a particle is given by \(x=20 t-5 t^{3}\), where \(x\) is in meters and \(t\) is in seconds, when, if ever, is the particle's velocity ze
View solution Problem 22
The position of a particle moving along the \(x\) axis depends on the time according to the equation \(x=c t^{2}-b t^{3},\) where \(x\) is in meters and \(t\) i
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