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Sunee is training for an upcoming 5.0-km race. She starts out her training run by moving at a constant

pace of 4.3 m/s for 19 min. Then she accelerates at a constant rate until she crosses the finish line, 19.4

s later. What is her acceleration during the last portion of the training run?

23

W

Respuesta :

adioabiola

Here, we are required to determine Sunee's acceleration during the last portion of the training run.

Sunee's acceleration during the last portion of the training run is a = 0.0775m/s².

She starts out at a pace of 4.3 m/s for 19mins.

Therefore, after 19 mins = 1140seconds;

Sunee must have travelled a distance,

D = speed × time

D = 4.3 × 1140

D = 4902 m

However, the race is a 5km race, therefore the rest of the race, S = 5000 - 4902 = 98m

By using the equation of motion thus;

S = ut + (1/2)at².

  • where t = 19.4s
  • u = 4.3m/s
  • S = 98m
  • a = ?

Therefore;

  • 98 = 4.3(19.4) + (1/2) × a × (19.4)²

  • 98 = 83.42 + 188.18a

  • 14.58 = 188.18a

  • a = 14.58/188.18

a = 0.0775 m/s².

Therefore, Sunee's acceleration during the last portion of the training run is a = 0.0775m/s².

Read more:

https://brainly.com/question/24527971

hamzaahmeds

This question can be solved using the equations of motion.

The acceleration of Sunee during the last portion is "0.077 m/s²".

First, we will find the distance covered by Sunee in the first portion of uniform motion. So, we will use the equation for uniform motion here:

[tex]d = vt\\[/tex]

where,

d = distance covered = ?

v = uniform speed = 4.3 m/s

t = time interval = 19 min = 1140 s

Therefore,

[tex]d = (4.3\ m/s)(1140\ s)\\d = 4902\ m[/tex]

Now, the distance covered during the last portion will be:

s = Total Distance - d

s = 5 km - 4902 m = 5000 m - 4902 m

s = 98 m

Now, we use the second equation of motion to find out the acceleration during the last portion:

[tex]s = v_it+\frac{1}{2}at^2[/tex]

where,

s = 98 m

vi = initial speed = 4.3 m/s

t = time interval = 19.4 s

a = acceleration = ?

Therefore,

[tex]98\ m = (4.3\ m/s)(19.4\ s)+\frac{1}{2}a(19.4\ s)^2\\\\a = \frac{2(14.58\ m)}{376.36\ s^2}[/tex]

a = 0.077 m/s²

Learn more about the equations of motion here:

https://brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion.

Ver imagen hamzaahmeds

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