Question

An Olympic class sprinter starts a race from rest with an acceleration of 4.8 m/s2. Take her direction of motion as the positive direction.

Answer 1

The question is incomplete, the complete form is:

An Olympic class sprinter starts a race from rest with an acceleration of
`4.8 m/s^(2)`. Take her direction of motion as the positive direction.

(a) What is her speed 2.40 s later?

(b) Sketch a graph of her position vs. time for this period.

The answer is:

a) What is her speed 2.40 s later?

Her speed is 11.52 m/s

Step-by-step explanation:

The speed can be determined by means of the kinematic equation that corresponds to a Uniformly Accelerated Rectilinear Motion.

`v = v_(0) + at` (1)

Where
`v_(0)` is the initial velocity, a is the acceleration and t is the time.

Since she starts the race from rest her initial speed will be zero (
`v_(0) = 0`)

(a) What is her speed 2.40 s later?

By means of equation 1 it is gotten:

`v = 0 m/s + (4.8 m/s^(2))(2.40s)`

`v = 11.52 m/s`

Hence, her velocity is 11.52 m/s

(b) Sketch a graph of her position vs. time for this period.

The position of that period can be known by means of the next equation:

`x = v_(0)t + (1)/(2)at^(2)` (2)

However, the initial speed is equal to zero, so equation 2 can be rewritten:

`x = (1)/(2)at^(2)` (3)

But the acceleration is equal to
`4.8m/s^(2)`

`x = (1)/(2)(4.8m/s^(2))t^(2)` (4)

Then, equation 4 can be used to sketch a graph of her position vs. time.

For t = 0 s

`x = (1)/(2)(4.8m/s^(2))(0)^(2)`

`x = 0`

For t = 1 s

`x = (1)/(2)(4.8m/s^(2))(1s)^(2)`

`x = 2.4m`

For t = 1.40 s

`x = (1)/(2)(4.8m/s^(2))(1.40s)^(2)`

`x = 4.70m`

For t = 2 s

`x = (1)/(2)(4.8m/s^(2))(2s)^(2)`

`x = 9.6m`

For t = 2.40 s

`x = (1)/(2)(4.8m/s^(2))(2.40s)^(2)`

`x = 13.82m`