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.
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