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Alex73 [517]
3 years ago
7

During the final 5 seconds of a race a cyclist increased her velocity from 4 m/s to 7 m/s . What was her average acceleration du

ring those last 5 seconds
Mathematics
1 answer:
GREYUIT [131]3 years ago
8 0

Answer:

The average acceleration of the cyclist was 0.6 m/s².

Step-by-step explanation:

Acceleration:

The rate change of velocity per unit time is call the acceleration of the object.

a=\frac{v-u}t

u= initial velocity

v= final velocity

t= time taken change of velocity.

During the final 5 seconds of a race cyclist increased her velocity from 4m/s to 7 m/s

Here v= 7 m/s and u=4 m/s t=5 seconds

\therefore a=\frac{7\ m/s-4 \ m/s}{5s}

     =\frac{3}{5} \ m/s^2

     =0.6 m/s²

The average acceleration of the cyclist was 0.6 m/s².

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A parabola with vertex (h, k) and a vertical axis of symmetry is modeled by the equation y - k = a(x - h)2. Determine the vertex
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4 years ago
Anika is hiking on a rectengular trail at the nation park there are four resting spots along the corners of the trail on the map
Drupady [299]

Answer:

The perimeter is 14 units

<em></em>

Step-by-step explanation:

Given

A = (-2,2)

B = (1,2)

C = (1,-2)

D = (-2,-2)

Required

Determine the perimeter

To do this, we simply calculate the distance between the sides of the rectangle i.e. AB, BC, CD and DA.

Distance (d) is calculated using:

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

For AB

A = (-2,2) --- (x_1,y_1)

B = (1,2) --- (x_2,y_2)

AB = \sqrt{(-2 - 1)^2 + (2 - 2)^2} = \sqrt{9} = 3

For BC

B = (1,2) --- (x_1,y_1)

C = (1,-2) --- (x_2,y_2)

BC = \sqrt{(1-1)^2 + (2--2)^2} = \sqrt{16} = 4

For CD

C = (1,-2) --- (x_1,y_1)

D = (-2,-2) --- (x_2,y_2)

CD = \sqrt{(1 - -2)^2 + (-2 -- 2)^2} = \sqrt{9} = 3

For DA

D = (-2,-2) --- (x_1,y_1)

A = (-2,2)  --- (x_2,y_2)

DA = \sqrt{(-2--2)^2 + (-2-2)^2} = \sqrt{16} = 4

So, the perimeter (P) is:

P = AB + BC + CD + DA

P = 3 +4 + 3 + 4

P = 14

<em>The perimeter is 14 units</em>

5 0
3 years ago
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