As A Baseball Is Thrown, It Goes From 00 To 400 Ms In 0
1as A Baseball Is Being Thrown It Goes From 00 To 400 Ms In 018
The assignment involves analyzing various physics problems related to kinematics, including calculations of acceleration, velocity, displacement, and understanding fundamental physical quantities. The specific tasks include determining acceleration in different scenarios, converting units, understanding the concept of acceleration in terms of g's, analyzing circular motion, and interpreting graph slopes. Additionally, the assignment explores the distinctions between speed and velocity as well as the nature of fundamental physical quantities.
Paper For Above instruction
Physics is a fundamental science that deals with understanding the behavior of matter and energy through various quantitative measurements and mathematical descriptions. Among its core branches, kinematics studies the motion of objects without considering the forces that cause this motion. The problems presented here encompass a broad scope of kinematic principles, particularly involving linear motion, circular motion, and basic velocity and acceleration calculations. They also relate to understanding the nature of physical quantities, their units, and their vector or scalar nature, as well as interpreting graphical representations of motion.
Problem 1: Acceleration of a Thrown Baseball
A baseball accelerates from 0 m/s to 40 m/s in 0.18 seconds. To determine the acceleration, we apply the basic formula of constant acceleration:
a = Δv / Δt
where Δv is the change in velocity and Δt is the time taken. Substituting the given values:
a = (40 m/s - 0 m/s) / 0.18 s ≈ 222.22 m/s²
This is a very high acceleration, typical for objects in rapid motion. To express this in terms of g, where 1 g ≈ 9.81 m/s², we divide the acceleration by 9.81:
a in g's ≈ 222.22 / 9.81 ≈ 22.66 g
Problem 2: Child on a Spinning Merry-Go-Round
The child's tangential acceleration (a) on a circular path is given by:
a = v² / r
where v = 5 m/s and r = 3.0 m:
a = (5)^2 / 3 = 25 / 3 ≈ 8.33 m/s²
This acceleration acts toward the center of the circle, reflecting the inward force required for circular motion.
Problem 3: Car on a Curved Track
The acceleration in uniform circular motion is computed similarly:
a = v² / r
where v = 78 m/s, r = 240 m:
a = (78)^2 / 240 ≈ 6084 / 240 ≈ 25.35 m/s²
The car experiences an acceleration of approximately 25.35 m/s² directed toward the center of the curve, critical for maintaining its circular trajectory.
Problem 4: Accelerating Train
Given the train starts from rest with a constant acceleration of 0.6 m/s²:
- (a) The velocity after 38 seconds is calculated by:
v = a t = 0.6 m/s² 38 s = 22.8 m/s
- (b) To find the total time to reach 43 m/s, rearranged as:
t = v / a = 43 / 0.6 ≈ 71.67 s
Problem 5: Dropped Rock
Assuming initial velocity is zero and acceleration due to gravity g ≈ 9.81 m/s²:
- (a) Velocity upon impact:
v = g t = 9.81 m/s² 5.9 s ≈ 57.8 m/s downward
The velocity is directed downward, and its magnitude is approximately 57.8 m/s.
- The average velocity during fall:
v_avg = (initial + final) / 2 = (0 + 57.8) / 2 ≈ 28.9 m/s downward
- The height of the bridge (displacement):
s = (1/2) g t² = 0.5 9.81 (5.9)^2 ≈ 171.0 m
Problem 6: Airplane Takeoff
From rest to 56 m/s in 9 seconds:
- (a) Acceleration:
a = Δv / Δt = 56 / 9 ≈ 6.22 m/s²
- (b) Speed after 4 seconds:
v = a t = 6.22 4 ≈ 24.88 m/s
- (c) Distance traveled until reaching 56 m/s:
s = (1/2) a t² = 0.5 6.22 9² ≈ 252.2 m
Problem 7: Sports Car Acceleration and Deceleration
Convert 0 to 100 mph to m/s:
100 mph ≈ 44.7 m/s
- (a) Acceleration:
a = Δv / Δt = 44.7 / 8 ≈ 5.59 m/s²
- (b) Braking from 26 m/s to stop in 5.7 s:
a = -Δv / Δt = -26 / 5.7 ≈ -4.56 m/s²
Problem 8: Jet's Average Speed
Distance = 1,353 miles, time = 2.7 hours. Converting miles to meters and hours to seconds:
1 mile ≈ 1609.34 meters; 2.7 hours = 2.700 * 3600 ≈ 9720 seconds
Distance in meters = 1353 * 1609.34 ≈ 2,177,063 meters
Time in seconds = 9720 s
Average speed = Distance / Time ≈ 2,177,063 / 9720 ≈ 224.15 m/s
Problem 9: Fundamental Physical Quantities
The fundamental quantities include mass, length (distance), and time. These are considered fundamental because they are base quantities from which other quantities are derived. Speed and area are derived quantities, with speed being length per time, and area being length squared. Among these, mass, distance, and time are scalars, whereas velocity, which involves speed and direction, is a vector.
Problem 10: Physical Quantities and Their Nature
Physical quantities in this chapter include displacement, velocity, acceleration, and speed. Displacement and velocity are vectors, characterized by magnitude and direction. Acceleration is a vector indicating the rate of change of velocity. Speed is a scalar, representing magnitude only, with no directional component. These quantities are derived from basic fundamental quantities such as distance (or displacement), time, and mass (for dynamic quantities).
Problem 11: Speed vs. Velocity
Speed measures how fast an object moves regardless of direction; velocity includes both speed and direction. For example, a car traveling around a circular track at a constant speed of 60 km/h has a constant speed, but its velocity changes continually because its direction is changing. In this case, the magnitude of velocity stays constant, but the velocity vector varies in direction, indicating a non-zero acceleration.
Problem 13: Slope of a Distance-Time Graph
The slope of a distance versus time graph physically represents the velocity of the object. A steeper slope indicates a higher speed, while a flat line indicates zero traversal (rest). The slope is calculated as the change in distance divided by the change in time, which directly corresponds to the average velocity over that interval.
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