Physics 101
Ch-4:motion
in two dimensions
6th Edition - 2021
0797973884
Mechanical engineer at JUST Mohammad Y Rawashdeh
, Chapter #4: motion in two dimensions
)الحركة في اتجاهين(
الشابت السابق نفس المواضيع إال أخر درس فقط بإختالف بسيط، ر ر
الشابت نفس ❖ مقدمة :هاذ
نحك سيارة عىل شارع مستقيم ،كنا ر
الشابت السابق كنا ن
بإتجاهي ر
الشابت هوه الحركة موضوع
ي
الشابت هاذ حنشتغل عالمحورين مع بعض. ر نشتغل عىل محور واحد فقط إما ( )Xأو ()Y
إش عىل ( )Yما إله دخل وكل .) Y ( ب دخل ه إل ما ) X ( عىل بصت إش كل : ر
الشابت ❖ مالحظة ألخر
ي ِ ي
ب (.)X
4.1 The Position, Velocity, and Acceleration Vectors.
1. position.
̂𝒋𝒀 ⃗ = 𝑿𝒊̂ +
𝒓
ن ر
❖ متجه من شابت ثالث لما شحت الموضع حكيت مثال نقطة ( )4,3ي
يعن 4عالسينات و3
عالصادات.
2. displacement.
𝒊𝒓 ⃗ = 𝒓 𝒇 −
𝒓∆ = 𝒕𝒏𝒆𝒎𝒆𝒄𝒂𝒍𝒑𝒔𝒊𝒅
𝒋 𝒊𝒀 𝒓𝒇 = 𝑿𝒇 𝒊 + 𝒀𝒇 𝒋 , 𝒓𝒊 = 𝑿𝒊 𝒊 +
𝒋 ) 𝒊𝒀 ⃗ = (𝑿𝒇 − 𝑿𝒊 ) 𝒊 + (𝒀𝒇 −
𝒓∆
𝒋 𝒚𝒓∆ ⃗ = (∆𝑿) 𝒊 + (∆𝒀) 𝒋 = ∆𝒓𝒙 𝒊 +
𝒓∆
3. velocity.
⃗
𝒓∆
=̅
𝑽 = 𝒚𝒕𝒊𝒄𝒐𝒍𝒆𝒗 𝒂𝒗𝒂 𝟐 𝒚𝑽 , 𝒂𝒗𝒂 𝒔𝒑𝒆𝒆𝒅 = |𝑽| = √𝑽𝒙 𝟐 +
𝒕∆
𝒓𝒅 𝒙𝒅 𝒚𝒅
= 𝒚𝒕𝒊𝒄𝒐𝒍𝒆𝒗 𝒔𝒐𝒖𝒆𝒏𝒂𝒕𝒏𝒂𝒕𝒔𝒏𝒊 ̂𝒋 ) ( = ( ) 𝒊̂ +
𝒕𝒅 𝒕𝒅 𝒕𝒅
4. acceleration.
𝑽∆
=̅
𝒂 = 𝒏𝒐𝒊𝒕𝒂𝒓𝒆𝒍𝒆𝒄𝒄𝒂 𝒂𝒗𝒂.
𝒕∆
𝒗𝒅 𝒙𝒗𝒅 𝒚𝒗𝒅
= 𝒏𝒐𝒊𝒕𝒂𝒓𝒆𝒍𝒆𝒄𝒄𝒂 𝒔𝒖𝒐𝒆𝒏𝒂𝒕𝒏𝒂𝒕𝒔𝒏𝒊 (= ( )𝒊+ 𝒋)
𝒕𝒅 𝒕𝒅 𝒕𝒅
Page 1 of 44
,Ex) an object moving on the xy-plane from the point (2,4) to (-3,7) in 3s:
1- What is the magnitude of its displacement (in meter)?
2- What its ava velocity and ava speed during this time interval?
Sol:
1-
𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 = ∆𝒓 = ∆𝒙𝒊 + ∆𝒚𝒋
∆𝑟 = (−3 − 2) 𝑖 + (7 − 4) 𝑗 = −5 𝑖 + 3 𝑗
|∆𝑟| = √(−5)2 + (3)2 = √36 = 6 𝑚
2-
𝚫𝒓 −5 3
̅=
𝑽 =( 𝑖 + 𝑗) 𝑚/𝑠
𝚫𝒕 3 3
𝒂𝒗𝒂 𝒔𝒑𝒆𝒆𝒅 = |𝑽| = √𝑽𝒙 𝟐 + 𝑽𝒚 𝟐
−5 2
|𝑉| = √( ) + 12 = 1.945 𝑚/𝑠
3
Ex) an object on the xy-plane its velocity changing from (-2i+j) m/s to (4i-3j) m/s
in 1s, what is the magnitude of its acceleration?
Sol:
(𝑽𝒙𝒇 − 𝑽𝒙𝒊 )𝒊 + (𝑽𝒚𝒇 − 𝑽𝒚𝒊 )𝒋
𝒂=
∆𝒕
(4 − (−2)) 𝑖 + (−3 − 1) 𝑗
𝑎= = (6 𝑖 − 4 𝑗) 𝑚/𝑠 2
1
|𝒂| = √𝟔𝟐 + (−𝟒)𝟐 = √𝟒𝟐 𝒎/𝒔𝟐
Page 2 of 44
, Ex) the position vector of an object moving on the xy-plan is given as a function
of time (𝑟(𝑡) = (1 + 2𝑡 2 ) 𝑖 + (3 − 𝑡 3 ) 𝑗 𝑚), find:
1) the position at (t=2s).
Sol:
𝑟(2) = (1 + (2 × 4)) 𝑖 + (3 − 8) 𝑗 = (9 𝑖 − 5 𝑗)𝑚
2) the displacement from (t=0 s to t=2 s).
Sol:
𝑟(0) = (1 𝑖 + 3 𝑗)𝑚 , 𝑟(2) = (9 𝑖 − 5 𝑗)𝑚
∆𝑟 = (9 − 1) 𝑖 + (−5 − 3) 𝑗 = (8 𝑖 − 8 𝑗) 𝑚
3) the ava velocity and ava speed from (t=0 s to t=2 s).
Sol:
∆𝑟 8 −8
𝑉̅ = = ( ) 𝑖 + ( ) 𝑗 = (4 𝑖 − 4 𝑗)𝑚/𝑠
∆𝑡 2 2
𝑎𝑣𝑎 𝑠𝑝𝑒𝑒𝑑 = |𝑉| = √(4)2 + (−4)2 = √32 𝑚/𝑠
4) the instantaneous velocity at (t=2 s).
Sol:
𝒅𝒓 𝒅𝒙 𝒅𝒚
𝒊𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒖𝒐𝒔 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = = ( ) 𝒊̂ + ( ) 𝒋̂
𝒅𝒕 𝒅𝒕 𝒅𝒕
𝑑𝑟
𝑉(𝑡) = = (4𝑡) 𝑖 + (−3𝑡 2 ) 𝑗̂
𝑑𝑡
𝑉 (2) = (4 × 2) 𝑖 + (−3 × 22 ) 𝑗̂ = (8𝑖 − 12𝑗) 𝑚/𝑠
5) the ava acceleration from (t=0 s to t=1s)
Sol:
𝑉 (0) = 0 , 𝑉 (1) = (4 × 1) 𝑖 + (−3 × 12 ) 𝑗̂ = (4 𝑖 − 3 𝑗) 𝑚/𝑠
∆𝑉 (4 − 0) 𝑖 + (−3 − 0) 𝑗
𝑎̅ = = = (4 𝑖 − 3 𝑗) 𝑚/𝑠 2
∆𝑡 1
Page 3 of 44
Ch-4:motion
in two dimensions
6th Edition - 2021
0797973884
Mechanical engineer at JUST Mohammad Y Rawashdeh
, Chapter #4: motion in two dimensions
)الحركة في اتجاهين(
الشابت السابق نفس المواضيع إال أخر درس فقط بإختالف بسيط، ر ر
الشابت نفس ❖ مقدمة :هاذ
نحك سيارة عىل شارع مستقيم ،كنا ر
الشابت السابق كنا ن
بإتجاهي ر
الشابت هوه الحركة موضوع
ي
الشابت هاذ حنشتغل عالمحورين مع بعض. ر نشتغل عىل محور واحد فقط إما ( )Xأو ()Y
إش عىل ( )Yما إله دخل وكل .) Y ( ب دخل ه إل ما ) X ( عىل بصت إش كل : ر
الشابت ❖ مالحظة ألخر
ي ِ ي
ب (.)X
4.1 The Position, Velocity, and Acceleration Vectors.
1. position.
̂𝒋𝒀 ⃗ = 𝑿𝒊̂ +
𝒓
ن ر
❖ متجه من شابت ثالث لما شحت الموضع حكيت مثال نقطة ( )4,3ي
يعن 4عالسينات و3
عالصادات.
2. displacement.
𝒊𝒓 ⃗ = 𝒓 𝒇 −
𝒓∆ = 𝒕𝒏𝒆𝒎𝒆𝒄𝒂𝒍𝒑𝒔𝒊𝒅
𝒋 𝒊𝒀 𝒓𝒇 = 𝑿𝒇 𝒊 + 𝒀𝒇 𝒋 , 𝒓𝒊 = 𝑿𝒊 𝒊 +
𝒋 ) 𝒊𝒀 ⃗ = (𝑿𝒇 − 𝑿𝒊 ) 𝒊 + (𝒀𝒇 −
𝒓∆
𝒋 𝒚𝒓∆ ⃗ = (∆𝑿) 𝒊 + (∆𝒀) 𝒋 = ∆𝒓𝒙 𝒊 +
𝒓∆
3. velocity.
⃗
𝒓∆
=̅
𝑽 = 𝒚𝒕𝒊𝒄𝒐𝒍𝒆𝒗 𝒂𝒗𝒂 𝟐 𝒚𝑽 , 𝒂𝒗𝒂 𝒔𝒑𝒆𝒆𝒅 = |𝑽| = √𝑽𝒙 𝟐 +
𝒕∆
𝒓𝒅 𝒙𝒅 𝒚𝒅
= 𝒚𝒕𝒊𝒄𝒐𝒍𝒆𝒗 𝒔𝒐𝒖𝒆𝒏𝒂𝒕𝒏𝒂𝒕𝒔𝒏𝒊 ̂𝒋 ) ( = ( ) 𝒊̂ +
𝒕𝒅 𝒕𝒅 𝒕𝒅
4. acceleration.
𝑽∆
=̅
𝒂 = 𝒏𝒐𝒊𝒕𝒂𝒓𝒆𝒍𝒆𝒄𝒄𝒂 𝒂𝒗𝒂.
𝒕∆
𝒗𝒅 𝒙𝒗𝒅 𝒚𝒗𝒅
= 𝒏𝒐𝒊𝒕𝒂𝒓𝒆𝒍𝒆𝒄𝒄𝒂 𝒔𝒖𝒐𝒆𝒏𝒂𝒕𝒏𝒂𝒕𝒔𝒏𝒊 (= ( )𝒊+ 𝒋)
𝒕𝒅 𝒕𝒅 𝒕𝒅
Page 1 of 44
,Ex) an object moving on the xy-plane from the point (2,4) to (-3,7) in 3s:
1- What is the magnitude of its displacement (in meter)?
2- What its ava velocity and ava speed during this time interval?
Sol:
1-
𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 = ∆𝒓 = ∆𝒙𝒊 + ∆𝒚𝒋
∆𝑟 = (−3 − 2) 𝑖 + (7 − 4) 𝑗 = −5 𝑖 + 3 𝑗
|∆𝑟| = √(−5)2 + (3)2 = √36 = 6 𝑚
2-
𝚫𝒓 −5 3
̅=
𝑽 =( 𝑖 + 𝑗) 𝑚/𝑠
𝚫𝒕 3 3
𝒂𝒗𝒂 𝒔𝒑𝒆𝒆𝒅 = |𝑽| = √𝑽𝒙 𝟐 + 𝑽𝒚 𝟐
−5 2
|𝑉| = √( ) + 12 = 1.945 𝑚/𝑠
3
Ex) an object on the xy-plane its velocity changing from (-2i+j) m/s to (4i-3j) m/s
in 1s, what is the magnitude of its acceleration?
Sol:
(𝑽𝒙𝒇 − 𝑽𝒙𝒊 )𝒊 + (𝑽𝒚𝒇 − 𝑽𝒚𝒊 )𝒋
𝒂=
∆𝒕
(4 − (−2)) 𝑖 + (−3 − 1) 𝑗
𝑎= = (6 𝑖 − 4 𝑗) 𝑚/𝑠 2
1
|𝒂| = √𝟔𝟐 + (−𝟒)𝟐 = √𝟒𝟐 𝒎/𝒔𝟐
Page 2 of 44
, Ex) the position vector of an object moving on the xy-plan is given as a function
of time (𝑟(𝑡) = (1 + 2𝑡 2 ) 𝑖 + (3 − 𝑡 3 ) 𝑗 𝑚), find:
1) the position at (t=2s).
Sol:
𝑟(2) = (1 + (2 × 4)) 𝑖 + (3 − 8) 𝑗 = (9 𝑖 − 5 𝑗)𝑚
2) the displacement from (t=0 s to t=2 s).
Sol:
𝑟(0) = (1 𝑖 + 3 𝑗)𝑚 , 𝑟(2) = (9 𝑖 − 5 𝑗)𝑚
∆𝑟 = (9 − 1) 𝑖 + (−5 − 3) 𝑗 = (8 𝑖 − 8 𝑗) 𝑚
3) the ava velocity and ava speed from (t=0 s to t=2 s).
Sol:
∆𝑟 8 −8
𝑉̅ = = ( ) 𝑖 + ( ) 𝑗 = (4 𝑖 − 4 𝑗)𝑚/𝑠
∆𝑡 2 2
𝑎𝑣𝑎 𝑠𝑝𝑒𝑒𝑑 = |𝑉| = √(4)2 + (−4)2 = √32 𝑚/𝑠
4) the instantaneous velocity at (t=2 s).
Sol:
𝒅𝒓 𝒅𝒙 𝒅𝒚
𝒊𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒖𝒐𝒔 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = = ( ) 𝒊̂ + ( ) 𝒋̂
𝒅𝒕 𝒅𝒕 𝒅𝒕
𝑑𝑟
𝑉(𝑡) = = (4𝑡) 𝑖 + (−3𝑡 2 ) 𝑗̂
𝑑𝑡
𝑉 (2) = (4 × 2) 𝑖 + (−3 × 22 ) 𝑗̂ = (8𝑖 − 12𝑗) 𝑚/𝑠
5) the ava acceleration from (t=0 s to t=1s)
Sol:
𝑉 (0) = 0 , 𝑉 (1) = (4 × 1) 𝑖 + (−3 × 12 ) 𝑗̂ = (4 𝑖 − 3 𝑗) 𝑚/𝑠
∆𝑉 (4 − 0) 𝑖 + (−3 − 0) 𝑗
𝑎̅ = = = (4 𝑖 − 3 𝑗) 𝑚/𝑠 2
∆𝑡 1
Page 3 of 44
