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A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration of this sprinter and how long does it take him/her to reach that speed?

Respuesta :

carlosego

Answer

a = 3.674 m / s ^ 2


t = 3.13 s

Using the kinematic equations for the movement we have:


[tex]h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2[/tex] (1)


[tex]V_{f} = V_{0} + at[/tex] (2)


Where:


[tex]P_{0}[/tex] = initial position


[tex]V_{0}}[/tex] = initial velocity


a = acceleration


t = time in seconds


[tex]V_{f}[/tex] = final speed


We know:


[tex]P_{0}=0[/tex]


[tex]V_{0}= 0[/tex]

h = 18 m


[tex]V_{f} = 11.5\frac{m}{s}[/tex]

  So:

 From (2) we have that: [tex]t =\frac{V_{f}}{a}[/tex]


[tex]t =\frac{11.5}{a}[/tex]

From (1) we have to:


[tex]h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2[/tex]

Then we clear "a" to find the acceleration.


[tex]\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2[/tex]

Then, the time it takes to reach this speed is:


[tex]t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s[/tex]

wiseowl2018

Answer:

Part 1.

Given data:

  • v = final velocity  
  • u = initial velocity  
  • a = acceleration  
  • s = distance traveled  

For part 1.

v = 11.5m/s  

u = 0 which is the starting velocity  

a = ?  

s = 18.0m

Solution:

Using below formula:

v^2 = u^2 + 2as  

11.5² = 0 + 2a(18m)²  

a = 3.67 m / s^2

                                                                   a = 3.67 m / s^2

Part 2.

Given data:

v = 11.50m/s  

u = 0 which is the starting velocity  

a = 3.6795 m / s^2

Solution:

Formula:  v = u + at

11.5m/s = 0m/s + 3.6795 m / s^2 (t)

= t = 3.13 s

                                                                    t = 3.13 s