Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.

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Ch.7 Equations of Motion ( Ch.7 Equations of Motion ( 運運運 運運運 運運 運運 ) ) 1

description

3 symbol ( 符號 ) physical quantity ( 物理量 ) unit ( 單位 ) s displacement ( 位移 ) m u initial velocity ( 最初速度 ) ms -1 v final velocity ( 最後速度 ) ms -1 t time ( 時間 ) s a acceleration ( 加速度 ) ms -2

Transcript of Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.

Page 1: Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.

Ch.7 Equations of Motion (Ch.7 Equations of Motion ( 運動方程式運動方程式 ))

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7.1 Equations of Motion

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symbol (符號 )

physical quantity (物理量 )

unit (單位 )

s displacement (位移 ) m

u initial velocity (最初速度 ) ms-1

v final velocity (最後速度 ) ms-1

t time (時間 ) s

a acceleration (加速度 ) ms-2

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Class Exercise 7.1

A ball moves down a slope with an initial velocity of 5 ms-1. The acceleration of the ball is 2 ms-2. What is the velocity of the ball 3 s later?

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Class Exercise 7.1

A ball moves down a slope with an initial velocity of 5 ms-1. The acceleration of the ball is 2 ms-2. What is the velocity of the ball 3 s later?

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v = 5 + (2)(3) = 11 ms-1

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Class Exercise 7.2

A car starting from rest accelerates at 4.0 ms-2 for 10 s. Calculate the distance travelled by the car during this time.

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Class Exercise 7.2

A car starting from rest accelerates at 4.0 ms-2 for 10 s. Calculate the distance travelled by the car during this time.

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s = (0)(10) + (0.5)(4)(10)2 = 200 m

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Class Exercise 7.3 The driver of a lorry applies brakes to stop the lorry from initial velocity of 20 ms-1. The deceleration of the lorry is 5 ms-2. What is the stopping distance of the lorry?

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Class Exercise 7.3 The driver of a lorry applies brakes to stop the lorry from initial velocity of 20 ms-1. The deceleration of the lorry is 5 ms-2. What is the stopping distance of the lorry?

200

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(0)2 = (20)2 + (2)(-5)(s) 10s = 400 s = 40 m

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Class Exercise 7.4

Peter is running with uniform acceleration along a straight road. At time t = 0, his velocity is 1.2 ms-1. After 6 s, his velocity becomes 2.4 ms-1.

(a) Calculate his acceleration.

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Class Exercise 7.4

Peter is running with uniform acceleration along a straight road. At time t = 0, his velocity is 1.2 ms-1. After 6 s, his velocity becomes 2.4 ms-1.

(a) Calculate his acceleration.

u = 1.2 ms-1

t = 6 sv = 2.4 ms-1

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Class Exercise 7.4

Peter is running with uniform acceleration along a straight road. At time t = 0, his velocity is 1.2 ms-1. After 6 s, his velocity becomes 2.4 ms-1.

(a) Calculate his acceleration.

u = 1.2 ms-1

t = 6 sv = 2.4 ms-1

v = u + at2.4 = 1.2 + (a)(6) a = 0.2 ms-2

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(b) Calculate the distance traveled in this 6 s.

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(b) Calculate the distance traveled in this 6 s.

v2 = u2 + 2as(2.4)2 = (1.2)2 + 2(0.2)(s) s = 10.8 m

OR s = ut + (0.5)at2

= (1.2)(6) + (0.5)(0.2)(6)2

s = 10.8 m

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Class Exercise 7.5

A car is moving with uniform deceleration along a straight road. Its velocity decreases from 20 ms-1 to 10 ms-1` after travelling a distance of 50 m.

(a) What is its deceleration?

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Class Exercise 7.5

A car is moving with uniform deceleration along a straight road. Its velocity decreases from 20 ms-1 to 10 ms-1` after travelling a distance of 50 m.

(a) What is its deceleration?

u = 20 ms-1

s = 50 mv = 2.4 ms-1

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Class Exercise 7.5 A car is moving with uniform deceleration along a straight road. Its velocity decreases from 20 ms-1 to 10 ms-1` after travelling a distance of 50 m.

(a) What is its deceleration?

u = 20 ms-1

s = 50 mv = 2.4 ms-1

v2 = u2 + 2as(10)2 = (20)2 + 2(a)(50) a = -3 ms-2

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(b) How much further distance will the car travel before it comes to rest?

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(b) How much further distance will the car travel before it comes to rest?

v2 = u2 + 2as(0)2 = (10)2 + 2(-3)(s) 6s = 100 s = 16.7 m

u = 10 ms-1

a = -3 ms-2

v = 0

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Homework: Assignment Book 1. Chapter 7: equations of Motion Q.1 – Q.7 (p.26-33) 2. Deadline: 2009 - ___ - ____