Скачать или смотреть Batterjee Medical College General Physics Chapter 3 Motion at Constant Acceleration كلية البترجي

Motion at Constant Acceleration- horizontal motion
Imagine a moving object that can be modeled as a particle. If it begins from position 𝑥 0 and initial velocity 𝑣 0 and moves in a straight line with a constant acceleration 𝑎, its subsequent position and velocity are described by the following kinematic equations (assuming 𝑥 𝑖 = 0 ):
𝑣 = 𝑣 0 + 𝑎𝑡
𝑥 = 𝑣𝑡 + 1 𝑎𝑡 2
𝑣 2 = 𝑣 2 0 + 2 𝑎𝑥
You will gain a great deal of experience in the use of these equations by solving a number of exercises and problems. Many times you will discover that more than one method can be used to obtain a solution. Remember that these equations of kinematics cannot be used in a situation in which the acceleration varies with time. They can be used only when the acceleration is constant.
Example 1
A body starts from rest and accelerates at 3 𝑚/𝑠 2 for 4 s. Calculate (a) its final velocity, (b) the distance traveled.
Solution:
Considering 𝑣 0 = 0 (𝑠𝑡𝑎𝑟𝑡𝑠𝑓𝑟𝑜𝑚𝑟𝑒𝑠𝑡), 𝑎 = 3 𝑚/𝑠 2
(a) 𝑣 = ?
𝑣 = 𝑣 0 + 𝑎𝑡 𝑣 = 0 + 3 × 4 = 12 𝑚/𝑠
(b) 𝑥 = ?
𝑥 = 𝑣𝑡 + 1 𝑎𝑡 2
1 𝑥 = 0 × (3) + × 3 × (4) 2 = 24 𝑚
xample 2
A car moving with velocity 10 𝑚/𝑠 accelerates uniformly at 2 𝑚/𝑠 2 . Calculate its velocity after traveling 200 m.
Solution:
𝑣 0 = 10 𝑚/𝑠, 𝑎 = 2 𝑚/𝑠 2 , 𝑥 = 200 𝑚 , 𝑣 = ?
𝑣 2 = 𝑣 2 0 + 2 𝑎𝑥
𝑣 2 = (10) 2 + 2 × 2 × 200
𝑣 2 = 900
𝑣 = √900 = 30 𝑚/𝑠
Exercise 1
In an electron gun, an electron is accelerated uniformly from rest to a velocity of in a distance of 0.10 m. Calculate the acceleration.
The answer: 𝑎 = 8 × 10 15 𝑚/𝑠 2
Exercise 2
A car slows down from 28 m/s to rest in a distance of 88 m. What was its acceleration, assumed constant?
The answer: 𝑎 = − 4.5 𝑚/𝑠 2
Exercise 3
A car accelerates from 14 m/s to 21 m/s in 6 s. What was its acceleration? How far did it travel in this time? Assume constant acceleration.
𝑎 = 1.2 𝑚/𝑠 2 𝑎𝑛𝑑 𝑥 = 102 𝑚
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