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

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

a = 3.674 m / s ^ 2

t = 3.13 s

Using the kinematic equations for the movement we have:

`h(t) = P_(0) + Vot + (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(m)/(s)`

So:

From (2) we have that:
`t =(V_(f))/(a)`

`t =(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.

`(36)/(t^2) = a\\a = (36)/(((11.5)/(a))^2) \\\\a =(11.5^2)/(36)\\a = 3.674 m / s ^2`

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

`t =(V_(f))/(a)\\t =(11.5)/(3.674)\\t = 3.13 s`