Question

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?

Answer 1

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

Answer 2

Answer

a = 3.674 m / s ^ 2

t = 3.13 s

Using the kinematic equations for the movement we have:

$h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2$ (1)

$V_{f} = V_{0} + at$ (2)

Where:

$P_{0}$ = initial position

$V_{0}}$ = initial velocity

a = acceleration

t = time in seconds

$V_{f}$ = final speed

We know:

$P_{0}=0$

$V_{0}= 0$

h = 18 m

$V_{f} = 11.5\frac{m}{s}$

  So:

 From (2) we have that: $t =\frac{V_{f}}{a}$

$t =\frac{11.5}{a}$

From (1) we have to:

$h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2$

Then we clear "a" to find the acceleration.

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

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

$t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s$