contestada

The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach top speed, which he maintains for the remaining 55 m. (a) What is the average horizontal component of force exerted on his feet by the ground during acceleration? (b) What is the speed of the sprinter over the last 55 m of the race (i.e., his top speed)?

Respuesta :

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration [tex]a_1[/tex] for a total time [tex]t_1[/tex] During this part of the motion, he covers a distance equal to [tex]s_1 = 45 m[/tex], until he finally reaches a velocity of [tex]v_1 = u + a_1t_1[/tex]. We can use the following suvat equation:

[tex]s_1 = u t_1 + \frac{1}{2}a_1t_1^2[/tex]

which reduces to

[tex]s_1 = \frac{1}{2}a_1 t_1^2[/tex] (1)

since u = 0.

- In the second part, he continues with constant speed [tex]v_1 = a_1 t_1[/tex], covering a distance of [tex]d_2 = 55 m[/tex] in a time [tex]t_2[/tex]. This part of the motion is a uniform motion, so we can use the equation

[tex]s_2 = v_1 t_2 = a_1 t_1 t_2[/tex] (2)

We also know that the total time is 10.0 s, so

[tex]t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)[/tex]

Therefore substituting into the 2nd equation

[tex]s_2 = a_1 t_1 (10-t_1)[/tex]

From eq.(1) we find

[tex]a_1 = \frac{2s_1}{t_1^2}[/tex] (3)

And substituting into (2)

[tex]s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1[/tex]

Solving for t,

[tex]s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s[/tex]

So from (3) we find the acceleration in the first phase:

[tex]a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2[/tex]

And so the average force exerted on the sprinter is

[tex]F=ma=(66 kg)(2.34 m/s^2)=154.5 N[/tex]

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

[tex]v_1 = u +a_1 t_1[/tex]

where we have

u = 0

[tex]a_1  =2.34 m/s^2[/tex] is the acceleration

[tex]t_1 = 6.2 s[/tex] is the time of the first part

Solving the equation,

[tex]v_1 = 0 +(2.34)(6.2)=14.5 m/s[/tex]

Acceleration is the rate of change of velocity. The speed of the sprinter over the last 55 m of the race is 14.516 m/s.

What is acceleration?

Acceleration is given as the rate of change of velocity. It is given by the formula,

[tex]a = \dfrac{v-u}{t}[/tex]

Given to us

Total distance, s = 100 m

mass of the sprinter, m = 66 kg

The first half of the total distance, s₁ = 45

The second half of the total distance, s₂ = 55

Total Time, t = 10 s

A.) We know that the force applied by the sprinter on the ground is equal,

F = m x a

We know that according to the first equation of motion,

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

Using it for the first half of the sprint,

[tex]s_1 = (0)t_1 + \frac{1}{2}at_1^2\\[/tex]

[tex]s_1 =\frac{1}{2}at_1^2\\[/tex]

Acceleration,

[tex]a =\dfrac{2s_1}{t_1^2}[/tex]

Also, we know that for the second half of the sprint, the sprinter is running at a constant velocity,

[tex]s_2 = t_2 \times v\\\\ s_2 = t_2 \times a_1 \times t_1[/tex]

Total time can be written as,

[tex]t = t_1 + t_2\\10 = t_1 + t_2\\ t_2 = (10-t_1)[/tex]

Substitute the value [tex]t_2[/tex] in  [tex]s_2 = t_2 \times a_1 \times t_1[/tex],

[tex]s_2 = (10-t_1) \times a_1 \times t_1[/tex]

Substitute the value of acceleration,

[tex]s_2 = (10-t_1) \times(\dfrac{2s_1}{t_1^2}) \times t_1[/tex]

We know the value of s₁ and s₂,

[tex]55 = (10-t_1) \times(\dfrac{90}{t_1^2}) \times t_1[/tex]

[tex]55 = (10-t_1) \times(\dfrac{90}{t_1})\\\\55 = \dfrac{900}{t_1} -90\\\\t_1 = 6.2\rm \ sec[/tex]

B.)  The speed of the sprinter over the last 55 m of the race

Using the first equation of motion,

[tex]at = v-u\\\\(\dfrac{2s_1}{t_1^2})t_1 = v-0\\v = 14.516\ m/s[/tex]

Hence, the speed of the sprinter over the last 55 m of the race is 14.516 m/s.

Learn more about Acceleration:

https://brainly.com/question/12134554