CLASS 9
Observe the following velocity-time graph of a body in motion and answer.
i) Which part of the graph represents uniform motion of the body?
a)AB
b)B
(c)CD
(d)GA
Calculate the acceleration of the body from C toD
a) 3.5 m/s2
(b)3.75m/s2
(c)4m/s2(
d)3.25m/s2
The displacement of the body in the first 8 seconds of motion is:
a)85 m
(b)80 m
(c)90 m
(d)82 m
Which part of the graph represents maximum acceleration?
(a)A-B
b)B-C
(c)A-C
d)C-D
Answers
- Answer:
Combine terms: 12a + 26b -4b – 16a.
(a) 4a + 22b,
(b) -28a + 30b,
(c) -4a + 22b,
(d) 28a + 30b.
Solution:
12a + 26b -4b – 16a.
= 12a – 16a + 26b – 4b.
= -4a + 22b.
Answer: (c)
4. Simplify: (4 – 5) – (13 – 18 + 2).
(a) -1,
(b) –2,
(c) 1,
(d) 2.
Solution:
(4 – 5) – (13 – 18 + 2).
= -1-(13+2-18).
= -1-(15-18).
= -1-(-3).
= -1+3.
= 2.
Answer: (d)
5. What is |-26|?
(a) -26,
(b) 26,
(c) 0,
(d) 1
Solution:
|-26|
= 26.
Answer: (b)
6. Multiply: (x – 4)(x + 5)
(a) x2 + 5x - 20,
(b) x2 - 4x - 20,
(c) x2 - x - 20,
(d) x2 + x - 20.
Solution:
(x – 4)(x + 5).
= x(x + 5) -4(x + 5).
= x2 + 5x – 4x – 20.
= x2 + x - 20.
Answer: (d)
7. Factor: 5x2 – 15x – 20.
(a) 5(x-4)(x+1),
(b) -2(x-4)(x+5),
(c) -5(x+4)(x-1),
(d) 5(x+4)(x+1).
Solution:
5x2 – 15x – 20.
= 5(x2 – 3x – 4).
= 5(x2 – 4x + x – 4).
= 5{x(x - 4) +1(x - 4)}.
= 5(x-4)(x+1).
Answer: (a).
8. Factor: 3y(x – 3) -2(x – 3).
(a) (x – 3)(x – 3),
(b) (x – 3)2,
(c) (x – 3)(3y – 2),
(d) 3y(x – 3).
Solution:
3y(x – 3) -2(x – 3).
= (x – 3)(3y – 2).
Answer: (c).
9. Solve for x: 2x – y = (3/4)x + 6.
(a) (y + 6)/5,
(b) 4(y + 6)/5,
(c) (y + 6),
(d) 4(y - 6)/5.
Solution:
2x – y = (3/4)x + 6.
or, 2x - (3/4)x = y + 6.
or, (8x -3x)/4 = y + 6.
or, 5x/4 = y + 6.
or, 5x = 4(y + 6).
or, 5x = 4y + 24.
or, x = (4y + 24)/5.
Therefore, x = 4(y + 6)/5.
Answer: (b).
10. Simplify:(4x2 - 2x) - (-5x2 - 8x).
Solution:
(4x2 - 2x) - (-5x2 - 8x)
= 4x2 - 2x + 5x2 + 8x.
= 4x2 + 5x2 - 2x + 8x.
= 9x2 + 6x.
= 3x(3x + 2).
Answer: 3x(3x + 2)
11. Find the value of 3 + 2 • (8 – 3)
(a) 25,
(b) 13,
(c) 17,
(d) 24,
(e) 15.
Solution:
3 + 2 • (8 – 3)
= 3 + 2 (5)
= 3 + 2 × 5
= 3 + 10
= 13
Answer: (d)
12. Rice weighing 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?
Solution:
33/4 ÷ 4 pounds.
= (4 × 3 + 3)/4 ÷ 4 pounds.
= 15/4 ÷ 4 pounds.
= 15/4 × 1/4 pounds.
= 15/16 pounds.
Now we know that, 1 pound = 16 ounces.
Therefore, 15/16 pounds = 15/16 × 16 ounces.
= 15 ounces.
Answer: 15 ounces.
13. Factor: 16w3 – u4w3
Solution:
16w3 – u4w3.
= w3(16 – u4).
= w3(42 - ((u2)2).
= w3(4 + u2)(4 - u2).
= w3(4 + u2)(22 - u2).
= w3(4 + u2)(2 + u)(2 - u).
Answer: w3(4 + u2)(2 + u)(2 - u).
14. Factor: 3x4y3 – 48y3.
Solution:
3x4y3– 48y3.
= 3y3(x4 – 16).
= 3y3[(x2)2 - 42].
= 3y3(x2 + 4)(x2 - 4).
= 3y3(x2 + 4)(x2 - 22).
= 3y3(x2 + 4)(x + 2)(x -2).
Answer: 3y3(x2 + 4)(x + 2)(x -2)
Answer:
i) b) BC
ii) b) 3.75 m/s²
iii) b) 80m
iv) d) C-D
Explanation:
To answer this question, we need to be familiar with the three equations of motion.
1. v = u + at (First Equation of Motion)
2. S = ut + ½at² (Second Equation of Motion)
3. v² = u² + 2aS (Third Equation of Motion)
where v is the final velocity, u is the initial velocity, S is the displacement, a is the acceleration and t is time.
i) Uniform motion of the body means when the body is moving at the same speed/velocity. As we can see in the graph, that from 8s to 14s the body maintains its velocity at 15 m/s therefore that part (From B to C) depicts the uniform motion of the body. Option b is correct.
ii) To calculate the acceleration of the body from C to D, we can use the 1st equation of motion, that is v = u + at
At point C the velocity is 15 m/s so u = 15 m/s and at point D the velocity is 30 m/s so v = 30 m/s and the time spent between points C & D is from 14s to 18s that is 4s
Putting the values to solve the equation
v = u + at
30 = 15 + 4a
15 = 4a
a = 15/4
a = 3.75 m/s². Option b is correct.
iii) To calculate the displacement of the body in the first 8 seconds we need to first use the 1st equation of motion to calculate the acceleration and then use either of the 2nd or 3rd equations of motion to evaluate the Displacement.
In this case, u = 5 m/s, v = 15 m/s and t = 8s
Applying v = u + at, we get
15 = 5 + 8a
8a = 10
a = 5/4 m/s² = 1.25 m/s²
Now applying v² = u² + 2aS, we get
15² = 5² + 2 × 1.25 S
225 = 25 + 2.5S
2.5 S = 200
S = 200/2.5
S = 80m . Option b is correct.
iv) As we have already know the acceleration for all three parts from our previous solutions, we can just compare them to find out which part has the highest acceleration.
From A to B the value of acceleration was 1.25 m/s² .
From B to C the value of acceleration was 0 m/s² as the body maintained uniform Velocity.
From C to D the value of acceleration was 3.75 m/s².
As we can clearly see, the part with the highest acceleration was from C to D, therefore Option d is correct.
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