A world record was set for the men's 100-m dash in the 2008 olympic games in beijing by usain bolt of jamaica. bolt "coasted" across the finish line with a time of 9.69 s. if we assume that bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, (a) calculate his maximum speed and (b) his acceleration. ans: (a) 12.2 m/s, (b) 4.07 m/s2

Respuesta :

Usain Bolt's acceleration  is 4.07 m/s² for the first 3.00 s and his maximum velocity is 12.2 m/s.

Bolt  starts from rest and accelerates with an acceleration a for a time t₁ =3.00 s and at the end of this time, he reaches a speed v. He travels a distance s₁ during this time.

Therefore,

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

His final velocity at the end of time t₁ is given by,

[tex]v=at_1......(2)[/tex]

He travels a further distance s₂ with a constant velocity v for a time t₂. Therefore,

[tex]s_2=vt_2[/tex]

From equation (2),

[tex]s_2=vt_2=at_1t_2[/tex]......(3)

The total distance traveled is s =100 m and

[tex]s= s_1+s_2[/tex]

From equations (1) and (3)

[tex]s=s_1+s_2=\frac{1}{2} at_1^2+at_1t_2=at_1(\frac{t_1}{2} +t_2)[/tex]

write an expression for a.

[tex]a=\frac{s}{t_1(\frac{t_1}{2} +t_2)}[/tex]

The total time taken to complete the race is 9.69 s, therefore,

[tex]t_2=9.69s-3.00s=6.69 s[/tex]

Substitute 100 m for s, 3.00 s for t₁ and 6.69 s for  t₂ and calculate the acceleration.

[tex]a=\frac{s}{t_1(\frac{t_1}{2} +t_2)}\\ =\frac{100m}{(3.00s)(\frac{3.00 s}{2}+(6.69 s) } \\ =4.07m/s^2[/tex]

In equation (2) substitute 4.07 m/s²for a and 3.00 s for t₁ .

[tex]v=at_1=(4.07m/s^2)(3.00s)=12.2m/s[/tex]

The maximum velocity of Usain Bolt is 12.2 m/s and his acceleration is 4.07m/s².