Respuesta :
Answer:
The total time for the race is 11.6 seconds
Explanation:
The parameters given are;
Total distance ran by Willie = 100.0 m
Initial acceleration = 2.00m/s²
Top speed reached with initial acceleration = 12.0 m/s
Point where Willie start to fade and decelerate = 16.0 m from the finish line
Speed with which Willie crosses the finish line = 8.00 m/s
The time and distance covered with the initial acceleration are found using the following equations of motion;
v = u₀ + a·t
v² = u₀² + 2·a·s
Where:
v = Final velocity reached with the initial acceleration = 12.0 m/s
u₀ = Initial velocity at the start of the race = 0 m/s
t = Time during acceleration
a = Initial acceleration = 2.00 m/s²
s = Distance covered during the period of initial acceleration
From, v = u₀ + a·t, we have;
12 = 0 + 2×t
t = 12/2 = 6 seconds
From, v² = u₀² + 2·a·s, we have;
12² = 0² + 2×2×s
144 = 4×s
s = 144/4 =36 meters
Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;
Distance covered at top speed = 100 - 36 - 16 = 48 meters
Time, [tex]t_t[/tex] of running at top speed = Distance/velocity = 48/12 = 4 seconds
The deceleration from top speed to crossing the line is found as follows;
v₁² = u₁² + 2·a₁·s₁
Where:
u₁ = v = 12 m/s
v₁ = The speed with which Willie crosses the line = 8.00 m/s
s₁ = Distance covered during decelerating = 16.0 m
a₁ = Deceleration
From which we have;
8² = 12² + 2 × a × 16
64 = 144 + 32·a
64 - 144 = 32·a
32·a = -80
a = -80/32 = -2.5 m/s²
From, v₁ = u₁ + a₁·t₁
Where:
t₁ = Time of deceleration
We have;
8 = 12 + (-2.5)·t₁
t₁ = (8 - 12)/(-2.5) = 1.6 seconds
The total time = t + [tex]t_t[/tex] + t₁ =6 + 4 + 1.6 = 11.6 seconds.
Willie's total time to cover the race is 11.6 seconds.
What is a velocity-time graph?
A velocity-time graph is a graph that is used to represent a way in which the motion of an object can be represented with the velocity change on the vertical (y)-axis and change in time on the horizontal (x)-axis.
From the given information:
- Willie's total distance covered = 100.0 m
- Initial acceleration (u) when the race commence = 0 m/s²
- Final velocity reached = 12.0 m/s
Using the following first equation of motion:
- v = u + at (where u = 0; at rest)
The time during the first acceleration can be computed as:
v = u + at
12 = 0 + (2 × t)
t = 12/2
t = 6 seconds
Using the fourth equation of motion, the distance covered in the first acceleration is determined as follows:
[tex]\mathbf{v^2 = u^2 + 2as}[/tex]
[tex]\mathbf{12^2 = (0)^2 + 2(2) \times s}[/tex]
144 = 4s
s = 144/4
s = 36 meters
However, he maintained a speed of 12.0 m/s till he is 16.0 m away from the finish line when he starts to decelerate.
Hence, at his highest speed;
- The distance covered = 100 - 36 - 16 = 48 meters.
Recall that:
[tex]Time = \dfrac{distance}{velocity}[/tex]
Thus, the time for running at his highest speed is:
[tex]\mathbf{time = \dfrac{48 m}{12 m/s}}[/tex]
time = 4 seconds
His deceleration from his highest speed can be computed as:
[tex]\mathbf{v^2 = u^ 2 + 2as}[/tex]
here:
- v = 8.00 m/s
- u = 12.0 m/s
- s = h = 16.0 m
[tex]\mathbf{v^2 - u^2=2as}[/tex]
[tex]\mathbf{a = \dfrac{v^2 - u^2}{2s}}[/tex]
[tex]\mathbf{a = \dfrac{8^2 - 12^2}{2(16)}}[/tex]
[tex]\mathbf{a = \dfrac{64 - 144}{32}}[/tex]
a = -2.5 m/s²
Finally, using the first equation of motion, the total time can be computed as:
[tex]\mathbf{v_1 = u_1 + a_1t_1}[/tex]
[tex]\mathbf{8= 12 + (-2.5)t_1}[/tex]
[tex]\mathbf{t_1 = \dfrac{(8- 12)}{ (-2.5)}}[/tex]
[tex]\mathbf{t_1 = \dfrac{(-4)}{ (-2.5)}}[/tex]
[tex]\mathbf{t_1 = 1.6\ seconds}[/tex]
Thus, the total time = (6+4+1.6) seconds
The total time = 11.6 seconds
Learn more about the velocity-time graph here:
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