This belongs to Scalars and Vectors

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1

2. This is a REGULAR CLASS. Please make a
note by writing down what you have
seen here.
Make a separate note for CURRENT
ELECTRICITY.

3. Scalars & Vectors OLD is
GOLD Discussion

6. 𝐻𝑒𝑟𝑒 𝐴𝐵 ≠ 0 𝑡ℎ𝑒𝑛 𝑖𝑡 𝑚𝑢𝑠𝑡 𝑏𝑒
𝑐𝑜𝑠𝜃 = 0
→ 𝜃 = 90°

30. Numerical Problems

31. A truck travelling at a 50 km/h, north turns west at the same speed. What is the
change in velocity?
50 km/h
𝑽𝑨
𝑽𝑩
50 km/h
Change in velocity =∆𝑽 = 𝑽𝑩 − 𝑽𝑨
𝑽𝑩
𝑽𝑨
−𝑽𝑨
. . .. . . . . . . . . .
.
..
.
.
.
.
.
.
.
.
.
.
∆𝑉 = 𝑉𝐴
2 + 𝑉𝐵2
= 𝟓𝟎 𝟐 km/hr.
𝑽𝑨
𝑽𝑩
−𝑽𝑨
.
..
.
.
.
.
.
.
.
.
.
.
Alternative

32. Due East
Towards East
𝜃
North of East by 𝜃
E
N
S
W
North East
45°
45°
E
N
S
W
30°
N
E
S
W
E
N
S
W
N
𝜃
East of North by 𝜃
E
S
W E
N
S
W

33. A sail boat sails 2km east then 4km southwest, then an additional distance is unknown direction.
Its final position is 5 km directly east of starting point, Find the magnitude and direction of third
leg of journey.
Solution
Here
Rx = Ax + Bx + Cx
5 = 2 + (-4sin45°) + Cx
∴ 𝐶𝑥 = 5.83 𝑘𝑚
Ry = Ay + By + Cy
0 = O + (-4cos 45°) + Cy
∴ 𝐶𝑦 = 2.83 𝑘𝑚
Now
C = 𝒄𝒙
𝟐 + 𝑪𝒚𝟐
= 5.83 2 + 2.83 2
= 6.48 km
Direction : 𝐭𝐚𝐧
_𝟏 𝑪𝒚
𝑪𝒙
= 𝑇𝑎𝑛
_1 2.83
5.83
= 25.89° North of East
Alternatively
𝐴 = 2𝑖
𝐵 = −4𝑠𝑖𝑛45°𝑖 − 4𝑐𝑜𝑠45°𝑗 = −2.82𝑖 − 2.83𝑗
𝐶 = 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
𝑅 = 5𝑖
𝑺𝒊𝒏𝒄𝒆: 𝑹 = 𝑨 + 𝑩 + 𝑪
5𝑖 = 2𝑖 − 2.83𝑖 − 2.83𝑗 + 𝐶
𝐶 = 5 − 2 + 2.83 𝑖 + 2.83𝑗
𝐶 = 5.83𝑖 + 2.83𝑗
𝑁𝑜𝑤
𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 ∶ 𝑪 = 𝟓. 𝟖𝟑 𝟐 + 𝟐. 𝟖𝟑 𝟐 = 6.48 𝑘𝑚
𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛: 𝑻𝒂𝒏
_𝟏 𝑪𝒚
𝑪𝒙
= 𝑇𝑎𝑛
_1 2.83
5.83
=25.89° North of East

34. Of west

38. Ans:

39. X
Y

41. UDAY KHANAL
Department of Physics
CCRC
41

42. The essence of SCIENCE: ask an impertinent question, and
you are on the way to a pertinent answer.

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