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Kobotan [32]
3 years ago
5

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

Mathematics
2 answers:
DochEvi [55]3 years ago
7 0

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

Speed=\frac{distance}{time}

in this problem we have

distance=400\ m \\time=120\ sec

substitute in the formula

Speed=\frac{400}{120}

Speed=3.5\frac{m}{sec}

therefore

<u>the answer Part a) is</u>

the speed is equal to 3.5\frac{m}{sec}

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

magnitude=3.5\frac{m}{sec}

direction=North\ East\ (NE)

therefore

<u>the answer Part b) is</u>

the velocity is 3.5\frac{m}{sec}\ North\ East\ (NE)

Part c)

we know that

the acceleration is equal to the formula

a=\frac{V2-V1}{t2-t1}

in this problem we have

V2=0 \\V1=3.5\frac{m}{sec}

t2=15\ sec\\t1=0

substitute in the formula

a=\frac{0-3.5}{15-0}

a=-\frac{3.5}{15}\frac{m}{sec^{2}}

a=-\frac{7}{30}\frac{m}{sec^{2}}

therefore

<u>the answer Part c) is</u>

the acceleration is -\frac{7}{30}\frac{m}{sec^{2}}

This is an example of negative acceleration

steposvetlana [31]3 years ago
6 0

Step-by-step explanation:

1. It is given that,

Distance covered by the cyclist, d = 400 m

Time taken, t = 120 s

Speed = distance / time taken

v=\dfrac{400\ m}{120\ s}

v = 3.33 m/s

So, the speed of the cyclist is 3.33 m/s.

2. The cyclist comes to a stop position within 15 seconds. Its acceleration is given by :

a=\dfrac{v-u}{t}

a=\dfrac{0-3.33\ m/s}{15\ s}  

a=-0.22\ m/s^2

So, the acceleration of the cyclist is -0.22\ m/s^2. It is an example of negative acceleration as the cyclist is decelerating. Hence, this is the required solution.              

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Joshua sells a pack of pens for $3.15, which is 5 percent more than he pays for them. Which equation will help find x, the amoun
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Answer:

x+0.05x=3.15 which has one solution

Step-by-step explanation:

Complete question below:

Joshua sells a pack of pens for 3.15 which is 5 percent more than he pays for them. which equation will help find x, the amount he pays for a pack of pens? how many solutions will this equation have? x+0.05x=3.15, which has infinitely many solutions x+0.5x=3.15, which has one solution x+0.05×=3.15 which has one solution 0.05×=3.15, which has no solution

Let

x= the amount Joshua pays for a pack of pen

5% of x more than he paid for them

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0.05x + x = 3.15

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3 years ago
(i) 3 (a + 5b) + a (a + 4) (ii) y (10 - y) + 3 (y - 2) (iii) 2 (8a - 5b) + 3 (5a - 12) (iv) 3 (y - 3) + ( 8 - 6y + x) (v) a (a -
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Answer:

3 (a + 5b) + a (a + 4)  = a^2 + 7a + 15b

y (10 - y) + 3 (y - 2)  = - y^2+13y - 6

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3 (y - 3) + ( 8 - 6y + x) = -3y + x-1

a (a - 2b) + b (b + 2a - c) =a^2  + b^2  - bc

5 (x - y + z) + (4x + 3y) =9x - 2y +5z

Step-by-step explanation:

Solving (i):

3 (a + 5b) + a (a + 4)

Open brackets

3 (a + 5b) + a (a + 4)  = 3a + 15b + a^2 + 4a

Collect like terms

3 (a + 5b) + a (a + 4)  = a^2 + 4a+3a + 15b

3 (a + 5b) + a (a + 4)  = a^2 + 7a + 15b

Solving (ii)

y (10 - y) + 3 (y - 2)

Open bracket

y (10 - y) + 3 (y - 2)  = 10y - y^2 + 3y - 6

Collect like terms

y (10 - y) + 3 (y - 2)  = - y^2+10y  + 3y - 6

y (10 - y) + 3 (y - 2)  = - y^2+13y - 6

Solving (iii)

2 (8a - 5b) + 3 (5a - 12)

Open bracket

2 (8a - 5b) + 3 (5a - 12)  = 16a -10b + 15a - 36

Collect like terms

2 (8a - 5b) + 3 (5a - 12)  = 16a+ 15a -10b  - 36

2 (8a - 5b) + 3 (5a - 12)  = 31a -10b  - 36

Solving (iv)

3 (y - 3) + ( 8 - 6y + x)

Open bracket

3 (y - 3) + ( 8 - 6y + x) = 3y - 9 + 8 - 6y + x

Collect like terms

3 (y - 3) + ( 8 - 6y + x) = 3y  - 6y + x- 9 + 8

3 (y - 3) + ( 8 - 6y + x) = -3y + x-1

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a (a - 2b) + b (b + 2a - c)

Open bracket

a (a - 2b) + b (b + 2a - c) =a^2 - 2ab + b^2 + 2ab - bc

Collect like terms

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5 (x - y + z) + (4x + 3y)

Open brackets

5 (x - y + z) + (4x + 3y) =5x - 5y +5z+4x +3y

Collect like terms

5 (x - y + z) + (4x + 3y) =5x +4x +3y- 5y +5z

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