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Punjab Board 2027 Model Papers | Class 10 Mathematics Subjective Paper | Units 5 & 6
Class 10 Math Chapter 5 & 6 Grand Test
Punjab Board 2027 Model Papers for Class 10 Mathematics are designed according to the new board examination pattern. This subjective model paper covers Unit 5: Algebraic Fractions and Unit 6: Vectors in Plane, including important short questions, algebraic fractions, reciprocal equations, vectors, translation vectors, projectile motion, and long-answer questions. Students preparing for Punjab Board Class 10 Mathematics 2027 can use this Class 10 Math Model Paper 2027 for effective exam practice and board preparation.
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Board PAPER Pattern
03
MATHEMATICS
CHAPTER
WISE
WISE
Roll No. (in Figures):
(in Words):
Unit 5: Algebraic Fractions
Unit 6: Vectors in Plane
Unit 6: Vectors in Plane
Time : 20 Minutes
Marks: 15
Marks: 15
OBJECTIVE TYPE
| A | B | C | D | A | B | C | D | A | B | C | D |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 A | B | C | D | 6 A | B | C | D | 11 A | B | C | D |
| 2 A | B | C | D | 7 A | B | C | D | 12 A | B | C | D |
| 3 A | B | C | D | 8 A | B | C | D | 13 A | B | C | D |
| 4 A | B | C | D | 9 A | B | C | D | 14 A | B | C | D |
| 5 A | B | C | D | 10 A | B | C | D | 15 A | B | C | D |
Note:
Four possible answers A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of
that question with Marker or Pen ink. Cutting or filling two or more
circles will result in zero mark in that question.
Q1. 15
1.
Lowest form of
15xyz10
is:
Answer: (A)
Step 1: Find the HCF of the numerical parts 15 and 10. HCF = 5.
Step 2: The denominator (10) has no variables x, y or z, so no variable can be cancelled.
Step 3: Divide numerator and denominator by 5:
15xyz ÷ 510 ÷ 5
=
3xyz2
Step 4: 3xyz and 2 have no common factor, so this is the lowest form.
2.
(x3 − y3)
÷ (x − y) in simplified form is:
Answer: (B)
Step 1: Use the difference of cubes formula:
x3 − y3 = (x − y)(x2 + xy + y2)
Step 2: Write the division as a fraction:
(x − y)(x2 + xy + y2)x − y
Step 3: Cancel the common factor (x − y):
x2 + xy + y2
Check: Put x = 2, y = 1: (8 − 1) ÷ 1 = 7 and 4 + 2 + 1 = 7 ✓
3.
An equation of the form
5x + 64·5−x − 20 = 0
is called:
Answer: (B)
Step 1: Look at where the unknown x appears. It is in the exponent (power) of 5, in 5x and 5−x.
Step 2: An equation in which the variable is in the exponent is called an exponential equation.
Step 3: It is not radical (no variable under a root sign), not reciprocal (not of the form a·xn + b/xn…), and not fractional (x is not in a denominator as a base).
4.
Linear factors of
3x2 + 10x + 3 = 0
are:
Answer: (A)
Step 1: Multiply the first and last coefficients: 3 × 3 = 9.
Step 2: Find two numbers whose product is 9 and sum is 10 (the middle term): 9 and 1.
Step 3: Split the middle term: 3x2 + 9x + x + 3
Step 4: Group: 3x(x + 3) + 1(x + 3)
Step 5: Take (x + 3) common: (x + 3)(3x + 1)
Check: (x + 3)(3x + 1) = 3x2 + x + 9x + 3 = 3x2 + 10x + 3 ✓
5.
x + 5x + 2
is a:
Answer: (D)
Step 1: A rational expression is a ratio p(x)q(x) of two polynomials with q(x) ≠ 0.
Step 2: Here p(x) = x + 5 and q(x) = x + 2 are both polynomials, so it is a rational expression.
Step 3: It is not a polynomial (variable in denominator), not an equation (no “=” sign) and not a surd (no root).
6.
Which value of x makes
x − 5x2 − 25
undefined?
Answer: (C)
Step 1: A fraction is undefined when its denominator equals zero.
Step 2: Put the denominator equal to 0: x2 − 25 = 0
Step 3: Factorize: (x − 5)(x + 5) = 0
Step 4: So x − 5 = 0 or x + 5 = 0, giving x = 5 or x = −5.
Step 5: Both values make the denominator zero, so the expression is undefined at 5 and −5.
7.
What is the simplified form of
(a3 + b3)
÷ (a + b)?
Answer: (B)
Step 1: Use the sum of cubes formula:
a3 + b3 = (a + b)(a2 − ab + b2)
Step 2: Divide:
(a + b)(a2 − ab + b2)a + b
Step 3: Cancel (a + b): a2 − ab + b2
Check: Put a = 2, b = 1: (8 + 1) ÷ 3 = 3 and 4 − 2 + 1 = 3 ✓
8.
A vector having magnitude 1, is called:
Answer: (C)
Step 1: By definition, a vector whose magnitude (length) is 1 is a unit vector. Examples: i and j.
Step 2: The unit vector along any vector v is v ÷ |v|.
Step 3: A zero vector has magnitude 0, not 1, so (D) is wrong.
9.
If a = λb, then
a and b are:
Answer: (B)
Step 1: a = λb means a is a scalar multiple of b.
Step 2: Multiplying a vector by a scalar changes only its length (and reverses direction if λ is negative); it stays along the same line.
Step 3: Hence a and b are parallel vectors.
10.
The position vector of point P(3, −2) with respect to O is:
Answer: (B)
Step 1: The position vector of P(x, y) w.r.t. origin O is OP = xi + yj.
Step 2: Here x = 3 and y = −2.
Step 3: OP = 3i + (−2)j = 3i − 2j.
11.
Vector from point P(3,4) to origin is:
Answer: (C)
Step 1: Vector from P to Q = (coordinates of Q) − (coordinates of P).
Step 2: Here Q is the origin O(0, 0) and P is (3, 4).
Step 3: PO = (0 − 3, 0 − 4) = (−3, −4).
Step 4: In component form: PO = −3i − 4j.
12.
The intersecting point of x-axis and y-axis is:
Answer: (A)
Step 1: Every point on the x-axis has y = 0.
Step 2: Every point on the y-axis has x = 0.
Step 3: The point common to both has x = 0 and y = 0, i.e. the origin (0, 0).
13.
Two or more vectors lying in the same plane are called:
Answer: (B)
Step 1: “Co-” means together and “planar” refers to a plane.
Step 2: Vectors that lie in the same plane are therefore called coplanar vectors.
Step 3: Equal vectors have the same magnitude and direction; parallel vectors have the same or opposite direction — neither describes “same plane”.
14.
The new vertex of point A(−4,2) by a translation vector
7i + 2j is:
Answer: (B)
Step 1: Translation vector 7i + 2j = (7, 2): move 7 units right and 2 units up.
Step 2: New point A′ = A + translation vector.
Step 3: A′ = (−4 + 7, 2 + 2) = (3, 4).
15.
Equation for horizontal motion is:
Answer: (A)
Step 1: In horizontal motion there is no acceleration, so the horizontal velocity vx stays constant.
Step 2: Distance = velocity × time, so the horizontal distance is x = vxt.
Step 3: Options (C) and (D) use y (vertical distance), and (B) has a wrong negative sign, so they are incorrect.
Here is Subjective Part of Paper.
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Roll No. (in Figures):
(in Words):
Time : 2:10 Hours
Marks: 60
Marks: 60
Unit 5: Algebraic Fractions
Unit 6: Vectors in Plane
Unit 6: Vectors in Plane
SUBJECTIVE TYPE (PART - I)
Q2. Write short answers to any Six (6) questions:
(6 × 2 = 12)
(i) Reduce to the lowest form:
3a2 − 6ab2a2b − 4ab2
3a2 − 6ab2a2b − 4ab2
Answer: 32b
Step 1: Factorize the numerator: 3a2 − 6ab = 3a(a − 2b)
Step 2: Factorize the denominator: 2a2b − 4ab2 = 2ab(a − 2b)
Step 3: Write 3a(a − 2b)2ab(a − 2b) and cancel the common factors a and (a − 2b).
Step 4: Result = 32b
(ii) Reduce to the lowest form:
aca2x2 − ax
aca2x2 − ax
Answer: cx(ax − 1)
Step 1: Factorize the denominator: a2x2 − ax = ax(ax − 1)
Step 2: Write acax(ax − 1)
Step 3: Cancel the common factor a: cx(ax − 1)
(iii) Simplify:
a3a − b − b3b − a
Answer: a3 + b3a − b
Step 1: Make the denominators the same: b − a = −(a − b)
Step 2: −b3b − a = −b3−(a − b) = +b3a − b
Step 3: Add: a3a − b + b3a − b = a3 + b3a − b
Step 4: Using the sum of cubes, this equals (a + b)(a2 − ab + b2)a − b. No common factor remains.
(iv) Simplify:
4 − x2x − x + 2x + 5
Answer: 2 + 4x − x2x
Step 1: Combine the first two fractions (same denominator x):
4 − x2 − (x + 2)x = 4 − x2 − x − 2x = 2 − x − x2x
Step 2: Write 5 as 5xx.
Step 3: Add: 2 − x − x2 + 5xx = 2 + 4x − x2x
(v) Simplify:
24ℓm5ℓ + 10m × 5ℓ2 − 20m216mn
Answer: 3ℓ(ℓ − 2m)2n
Step 1: 5ℓ + 10m = 5(ℓ + 2m)
Step 2: 5ℓ2 − 20m2 = 5(ℓ2 − 4m2) = 5(ℓ − 2m)(ℓ + 2m)
Step 3: Expression = 24ℓm × 5(ℓ − 2m)(ℓ + 2m)5(ℓ + 2m) × 16mn
Step 4: Cancel 5, (ℓ + 2m) and m: 24ℓ(ℓ − 2m)16n
Step 5: Divide 24 and 16 by 8: 3ℓ(ℓ − 2m)2n
(vi) Simplify:
a2 − 4b2a2 + 2ba × 2a2 + 10aba2 + 3ab − 10b2
Answer: 2
Step 1: a2 − 4b2 = (a − 2b)(a + 2b)
Step 2: a2 + 2ab = a(a + 2b)
Step 3: 2a2 + 10ab = 2a(a + 5b)
Step 4: a2 + 3ab − 10b2 = (a + 5b)(a − 2b) (since 5 × (−2) = −10 and 5 − 2 = 3)
Step 5: Expression = (a − 2b)(a + 2b) × 2a(a + 5b)a(a + 2b) × (a + 5b)(a − 2b)
Step 6: Everything cancels except the 2. Result = 2
(vii) Reduce to the lowest form:
x2 − 7x + 12x2 − 6x + 9
Answer: x − 4x − 3
Step 1: x2 − 7x + 12 = x2 − 3x − 4x + 12 = (x − 3)(x − 4)
Step 2: x2 − 6x + 9 = (x − 3)2 = (x − 3)(x − 3)
Step 3: Cancel one (x − 3): x − 4x − 3
(viii) Reduce to the lowest form:
x2 + 3x − 187x − 21
Answer: x + 67
Step 1: x2 + 3x − 18 = x2 + 6x − 3x − 18 = (x + 6)(x − 3)
Step 2: 7x − 21 = 7(x − 3)
Step 3: Cancel (x − 3): x + 67
(ix) What value of x makes the fraction
7x − 5 undefined?
Answer: x = 5
Step 1: A fraction is undefined when its denominator is zero.
Step 2: Put x − 5 = 0, so x = 5.
Q3. Write short answers to any Six (6) questions:
(6 × 2 = 12)
(i) Reduce to the lowest form:
x2 + 5x + 6x2 − 4
Answer: x + 3x − 2
Step 1: x2 + 5x + 6 = x2 + 2x + 3x + 6 = (x + 2)(x + 3)
Step 2: x2 − 4 = (x − 2)(x + 2)
Step 3: Cancel (x + 2): x + 3x − 2
(ii) Simplify:
x5 + 5x
Answer: x2 + 255x
Step 1: LCM of the denominators 5 and x is 5x.
Step 2: x × x5 × x + 5 × 5x × 5 = x25x + 255x
Step 3: Add: x2 + 255x
(iii) Simplify:
x2 − 94 × 8x − 3
Answer: 2(x + 3)
Step 1: x2 − 9 = (x − 3)(x + 3)
Step 2: Expression = (x − 3)(x + 3) × 84 × (x − 3)
Step 3: Cancel (x − 3), and 8 ÷ 4 = 2.
Step 4: Result = 2(x + 3) = 2x + 6
(iv) Define "Reciprocal" equation.
Answer:
An equation which remains unchanged when x is replaced by 1x is called a reciprocal equation.
Example: 2x2 − 5x + 2 = 0. Replacing x by 1x and simplifying gives back the same equation.
(v) Find a vector triple in length of
3i − 2j, but opposite in direction.
Answer: −9i + 6j
Step 1: Triple in length means multiply by 3. Opposite direction means multiply by −1.
Step 2: Required vector = −3(3i − 2j)
Step 3: = −9i + 6j
(vi) Find two vectors that are double in magnitude of
3i − 5j,
one in the same direction of it and other in its opposite direction.
Answer: 6i − 10j and −6i + 10j
Step 1: Same direction, double magnitude: 2(3i − 5j) = 6i − 10j
Step 2: Opposite direction, double magnitude: −2(3i − 5j) = −6i + 10j
(vii) A ball is projected with velocity vector
v = 6i + 8j. What is the magnitude of velocity?
Answer: 10 units
Step 1: For v = xi + yj, |v| = √(x2 + y2)
Step 2: |v| = √(62 + 82) = √(36 + 64)
Step 3: = √100 = 10
(viii) Write the vector PQ in the form of
xi + yj, where P(−3, −6) and Q(1, 2).
Answer: PQ = 4i + 8j
Step 1: PQ = (coordinates of Q) − (coordinates of P)
Step 2: PQ = (1 − (−3), 2 − (−6)) = (1 + 3, 2 + 6) = (4, 8)
Step 3: PQ = 4i + 8j
(ix) If
u = 14i + 154j
then find |u|.
Answer: |u| = √2264 ≈ 3.76
Step 1: |u| = √( (14)2 + (154)2 )
Step 2: = √( 116 + 22516 ) = √22616
Step 3: = √2264 ≈ 3.76
Q4. Write short answers to any Six (6) questions:
(6 × 2 = 12)
(i) Check that the vectors
u = 8i + 4j and v = 2i + j
are parallel or not.
Answer: Parallel
Step 1: Check whether u is a scalar multiple of v: is 8i + 4j = k(2i + j)?
Step 2: Compare i-parts: 2k = 8, so k = 4. Compare j-parts: k = 4.
Step 3: Both give k = 4, so u = 4v. Hence u and v are parallel.
Note: the printed question had “v = 2i + i”; this was treated as a typing error for 2i + j.
(ii) If u = −5v, where
u = 10i − λj and v = 2i + 3j,
then find the value of λ.
Answer: λ = 15 (with v = −2i + 3j, see note)
Step 1: Put u = −5v: 10i − λj = −5(2i + 3j) = −10i − 15j
Step 2: Comparing i-parts gives 10 = −10, which is false. So with v = 2i + 3j no value of λ works.
Step 3: With v = −2i + 3j: −5v = 10i − 15j. Comparing j-parts: −λ = −15, so λ = 15.
Note: please check the original question. The i-parts only match if v = −2i + 3j.
(iii) If u = 5i − 4j and v = 6i − 3j,
then find u + v.
Answer: 11i − 7j
Step 1: Add i-parts: 5 + 6 = 11
Step 2: Add j-parts: −4 + (−3) = −7
Step 3: u + v = 11i − 7j
(iv) If a = 4i − j and b = −3i + 5j,
then find 3b + 4a.
Answer: 7i + 11j
Step 1: 3b = 3(−3i + 5j) = −9i + 15j
Step 2: 4a = 4(4i − j) = 16i − 4j
Step 3: 3b + 4a = (−9 + 16)i + (15 − 4)j = 7i + 11j
(v) If u = i − 2j, w = 3i + 3j
and u + 4v = w, then find v.
Answer: v = 12i + 54j
Step 1: From u + 4v = w, we get 4v = w − u.
Step 2: w − u = (3i + 3j) − (i − 2j) = 2i + 5j
Step 3: v = 14(2i + 5j) = 24i + 54j = 12i + 54j
(vi) If (xi + yj) − (2i − 4j) = −5i + 3j,
then find the values of x and y.
Answer: x = −3, y = −1
Step 1: Simplify the left side: (x − 2)i + (y + 4)j = −5i + 3j
Step 2: Compare i-parts: x − 2 = −5, so x = −3
Step 3: Compare j-parts: y + 4 = 3, so y = −1
(vii) If (x + 5)i + (y − 3)j = 2i + 7j,
then find the values of x and y.
Answer: x = −3, y = 10
Step 1: Equal vectors have equal components.
Step 2: x + 5 = 2, so x = −3
Step 3: y − 3 = 7, so y = 10
(viii) Define a translation vector.
Answer:
A translation vector is a vector that moves every point of a figure through the same distance in the same direction, without rotating it or changing its size or shape.
Example: Translation by ai + bj moves a point (x, y) to (x + a, y + b).
(ix) If AB = 4i − 5j and DC = 4i − yj
are two parallel opposite sides of a parallelogram, then what is the value of y?
Answer: y = 5
Step 1: Opposite sides of a parallelogram are equal and parallel, so AB = DC.
Step 2: 4i − 5j = 4i − yj
Step 3: Compare j-parts: −5 = −y, so y = 5.
(PART - II)
Note: Attempt any two (02) questions.
(2 × 8 = 16)
Q5.
(a) Simplify:
x − 2y3 + x + y6 − 2x − y15
4
(b) Simplify: 3a2 − 6ab6ab × 10b24ab − 8b2 4
Answer: 11x − 13y30
Step 1: LCM of 3, 6 and 15 = 30.
Step 2: Express with denominator 30:
10(x − 2y)30 + 5(x + y)30 − 2(2x − y)30
Step 3: Combine: 10(x − 2y) + 5(x + y) − 2(2x − y)30
Step 4: Expand: 10x − 20y + 5x + 5y − 4x + 2y30
Step 5: Collect like terms: x terms: 10x + 5x − 4x = 11x; y terms: −20y + 5y + 2y = −13y
Step 6: Result = 11x − 13y30
(b) Simplify: 3a2 − 6ab6ab × 10b24ab − 8b2 4
Answer: 54
Step 1: 3a2 − 6ab = 3a(a − 2b)
Step 2: 4ab − 8b2 = 4b(a − 2b)
Step 3: Expression = 3a(a − 2b)6ab × 10b24b(a − 2b) = 3a(a − 2b) × 10b26ab × 4b(a − 2b)
Step 4: Cancel (a − 2b): 30ab224ab2
Step 5: Cancel ab2 and divide 30, 24 by 6: 54
Q6.
(a)
Train is scheduled to cover a distance of 120 km at a certain average speed,
owing to a service checkup this average speed is reduced by 5 km/h,
and in consequence the journey takes 20 minutes more than the scheduled time.
What was the scheduled speed?
4
(b) One pipe fills in 5 hours, another in 8 hours. Second pipe closed after 2 hours. Find the total time. 4
Answer: 45 km/h
Step 1: Let the scheduled speed = x km/h. Scheduled time = 120x hours.
Step 2: New speed = (x − 5) km/h, new time = 120x − 5 hours.
Step 3: 20 minutes = 13 hour, so 120x − 5 = 120x + 13
Step 4: 120x − 5 − 120x = 13
Step 5: 120x − 120(x − 5)x(x − 5) = 13, i.e. 600x(x − 5) = 13
Step 6: Cross-multiply: x2 − 5x = 1800, so x2 − 5x − 1800 = 0
Step 7: Factorize: (x − 45)(x + 40) = 0, so x = 45 or x = −40
Step 8: Speed cannot be negative, so the scheduled speed = 45 km/h.
Check: 120 ÷ 45 = 2 h 40 min; 120 ÷ 40 = 3 h; difference = 20 min ✓
(b) One pipe fills in 5 hours, another in 8 hours. Second pipe closed after 2 hours. Find the total time. 4
Answer: 334 hours (3 hours 45 minutes)
Assumption: Both pipes start together to fill the tank; the second pipe is closed after 2 hours.
Step 1: Rate of 1st pipe = 15 per hour; rate of 2nd pipe = 18 per hour.
Step 2: Work done by both in 2 hours = 2(15 + 18) = 2 × 1340 = 1320
Step 3: Remaining work = 1 − 1320 = 720
Step 4: The first pipe alone does this: time = 7/201/5 = 720 × 5 = 74 hours
Step 5: Total time = 2 + 74 = 154 hours = 3 h 45 min.
Q7.
(a)
Rashid can paint a house in 6 days, but if he gets a helper he can do it in 4 days.
How long would it take the helper to paint the house alone?
4
(b) Plot A(−5, 3), B(−2, 3) and C(−4, 5) to form a triangle ABC. Also translate △ABC to △A′B′C′ by translation vector 5i − 2j. 4
Answer: 12 days
Step 1: Let the helper alone take x days. Rashid's one-day work = 16, helper's = 1x.
Step 2: Together they finish in 4 days, so their one-day work = 14.
Step 3: 16 + 1x = 14
Step 4: 1x = 14 − 16 = 3 − 212 = 112
Step 5: So x = 12 days.
(b) Plot A(−5, 3), B(−2, 3) and C(−4, 5) to form a triangle ABC. Also translate △ABC to △A′B′C′ by translation vector 5i − 2j. 4
Answer: A′(0, 1), B′(3, 1), C′(1, 3)
Step 1: Translation vector 5i − 2j means move 5 units right and 2 units down: (x, y) → (x + 5, y − 2).
Step 2: A′ = (−5 + 5, 3 − 2) = (0, 1)
Step 3: B′ = (−2 + 5, 3 − 2) = (3, 1)
Step 4: C′ = (−4 + 5, 5 − 2) = (1, 3)
Step 5: Plot A, B, C and join them to get △ABC. Plot A′, B′, C′ and join them to get △A′B′C′ (graph below).
(PART - III)
Note: Attempt any one (01) question.
(1 × 8 = 8)
Q8.
(a)
If 5i − 3j = m(i − 10j) + n(4i − 3j),
then find the value of m and n.
4
(b) Use vectors to show that triangle XYZ is an isosceles, where the points X, Y, Z have the coordinates (0, 0), (2, 0) and (1, 3) respectively. 4
Answer: m = −337, n = 4737
Step 1: Expand the right side: m(i − 10j) + n(4i − 3j) = (m + 4n)i + (−10m − 3n)j
Step 2: Compare i-parts: m + 4n = 5 …… (1)
Step 3: Compare j-parts: −10m − 3n = −3, i.e. 10m + 3n = 3 …… (2)
Step 4: From (1): m = 5 − 4n
Step 5: Put in (2): 10(5 − 4n) + 3n = 3, so 50 − 40n + 3n = 3, so −37n = −47, giving n = 4737
Step 6: m = 5 − 4 × 4737 = 185 − 18837 = −337
Check: 10m + 3n = −3037 + 14137 = 11137 = 3 ✓
(b) Use vectors to show that triangle XYZ is an isosceles, where the points X, Y, Z have the coordinates (0, 0), (2, 0) and (1, 3) respectively. 4
Answer: XZ = YZ, so △XYZ is isosceles.
Step 1: XY = Y − X = (2 − 0, 0 − 0) = 2i, so |XY| = 2
Step 2: YZ = Z − Y = (1 − 2, 3 − 0) = −i + 3j, so |YZ| = √((−1)2 + 32) = √10
Step 3: XZ = Z − X = (1 − 0, 3 − 0) = i + 3j, so |XZ| = √(12 + 32) = √10
Step 4: |XZ| = |YZ| = √10 ≠ |XY| = 2. Two sides are equal, so triangle XYZ is isosceles.
Q9.
(a)
A ball is projected with an initial velocity vector
v0 = 10i + 20j.
The horizontal component is in the x-direction and gravity is
g = 0i − 10j.
Find the maximum height and horizontal range.
4
(b) A car enters a loop with velocity vector v = 30i and exits with velocity v = 30j. What is the change in velocity vector? 4
Answer: Maximum height = 20 units, Range = 40 units
Step 1: Horizontal component vx = 10, vertical component vy = 20, and g = 10 (downward).
Step 2: At maximum height the vertical velocity is 0: 0 = 20 − 10t, so t = 2 s
Step 3: Maximum height: y = vyt − 12gt2 = 20(2) − 12(10)(2)2 = 40 − 20 = 20
Step 4: Total time of flight = 2 × 2 = 4 s
Step 5: Horizontal range: x = vxt = 10 × 4 = 40
(b) A car enters a loop with velocity vector v = 30i and exits with velocity v = 30j. What is the change in velocity vector? 4
Answer: Δv = −30i + 30j
Step 1: Change in velocity = final velocity − initial velocity.
Step 2: Δv = 30j − 30i = −30i + 30j
Step 3 (magnitude): |Δv| = √((−30)2 + 302) = √1800 = 30√2 ≈ 42.4
