Answer:
a) Vf = 10.8 [m/s]
b) See the image attached.
Explanation:
a)
To solve this part we must use the following kinematics equation:
[tex]v_{f} =v_{o} + (a*t)[/tex]
where:
Vf = final velocity [m/s^2]
Vo = initial velocity = 0
a = aceleration = 4.5 [m/s^2]
t = time = 2.4 [s]
Note: The initial velocity is taken as zero, as the athlete starts from rest
Vf = 0 + (4.5*2.4)
Vf = 10.8 [m/s]
b)
Now knowing that acceleration remains constant between the time interval between 0 to 2.4 seconds. We will take 5 different times between the 2.4 elapsed.
Dt = 2.4/5
Dt = 0.48 [s]
So the first time is 0, the second will be (0 + 0.48) = 0.48, the third (0.48 + 0.48) = 0.96, the fourth (0.96 + 0.48) = 1.44, and the fifth (1.44 + 0,48) = 1.92, and the last one (1.92 + 0.48) = 2.4
We need to find the velocity for each time calculated before, with the ecuation used in a) we cand calculate the different velocities.
Vfo = 0 + (4.5*0) = 0
Vf1 = 0 + (4.5*0.48) = 2.16 [m/s]
Vf2 = 0 + (4.5*0.96) = 4.32 [m/s]
Vf3 = 0 + (4.5*1.44) = 6.48 [m/s]
Vf4 = 0 + (4.5*1.92) = 8.64 [m/s]
Vf5 = 0 + (4.5*2.4) = 10.8 [m/s]
The next step is to use the following equation of kinematics, to be able to determine the position of the athlete for each speed.
[tex]v_{f}^{2} =v_{i}^{2} +(2*a*x)\\if \\v_{i}=0\\then\\x = \frac{v_{f}^{2}}{2*a}[/tex]
x1 = 0
x2 = (2.16)^2/(2*4.5) = 0.518 [m]
x3 = (4.32)^2/(2*4.5) = 2.074 [m]
x4 = (6.48)^2/(2*4.5) = 4.67 [m]
x5 = (8.64)^2/(2*4.5) = 8.29 [m]
x6 = (10.8)^2/(2*4.5) = 12.96 [m]
Now in the x-axis we put the time values found before as well the x values en the y-axis.
