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When a 1984 Alfa Romeo Spider sports cai” accelerates at
Chapter 2, Problem 64P(choose chapter or problem)
When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first \(20 \mathrm{~s}\) is extremely well modeled by the simple equation
\(v_{x}^{2}=\frac{2 P}{m} t\)
where \(P=3.6 \times 10^{4}\) watts is the car's power output, \(m=1200 \mathrm{~kg}\) is its mass, and \(v_{x}\) is in \(m / s\).
That is, the square of the car's velocity increases linearly with time.
a. What is the car's speed at \(t=10 \mathrm{~s}\) and at \(t=20 \mathrm{~s}\)?
b. Find an algebraic expression in terms of \(P, m\), and \(t\), for the car's acceleration at time \(t\)
c. Evaluate the acceleration at \(t=1 \mathrm{~s}$ and \(t=10 \mathrm{~s}\).
d. This simple model fails for \(t\) less than about \(0.5 \mathrm{~s}\). Explain how you can recognize the failure.
Equation Transcription:
Text Transcription:
20 s
v_x^2=2P/m t
P=3.6 x 10^4watts
m=1200 kg
v_x
m/s
t=1 s
t=10 s
t
0.5 s
Questions & Answers
QUESTION:
When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first \(20 \mathrm{~s}\) is extremely well modeled by the simple equation
\(v_{x}^{2}=\frac{2 P}{m} t\)
where \(P=3.6 \times 10^{4}\) watts is the car's power output, \(m=1200 \mathrm{~kg}\) is its mass, and \(v_{x}\) is in \(m / s\).
That is, the square of the car's velocity increases linearly with time.
a. What is the car's speed at \(t=10 \mathrm{~s}\) and at \(t=20 \mathrm{~s}\)?
b. Find an algebraic expression in terms of \(P, m\), and \(t\), for the car's acceleration at time \(t\)
c. Evaluate the acceleration at \(t=1 \mathrm{~s}$ and \(t=10 \mathrm{~s}\).
d. This simple model fails for \(t\) less than about \(0.5 \mathrm{~s}\). Explain how you can recognize the failure.
Equation Transcription:
Text Transcription:
20 s
v_x^2=2P/m t
P=3.6 x 10^4watts
m=1200 kg
v_x
m/s
t=1 s
t=10 s
t
0.5 s
ANSWER:
Step 1 of 4
(a)
We have to find the car's speed at \(t=10 \mathrm{~s}\) and \(t=20 \mathrm{~s}\).
The speed of the car can be calculated using the equation.
\(v_{x}^{2}=\frac{2 P}{m} t\)
Where,
\(P=3.6 \times 10^{4} \text { watts }\)
\(m=1200 \mathrm{~kg}\)
Thus, the speed of the car at \(t=10 \mathrm{~s}\) and \(t=20 \mathrm{~s}\) are given by
\(v_{x}(t=10 \mathrm{~s}) =\sqrt{\frac{2 P}{m}} t\)
\(=\sqrt{\frac{2 \times 3.6 \times 10^{4}}{1200}(10)}\)
\(=24.5 \mathrm{~m} / \mathrm{s}\)
\(v_{x}(t=20 \mathrm{~s})=\sqrt{\frac{2 P}{m}} t\)
\(=\sqrt{\frac{2 \times 3.6 \times 10^{4}}{1200}(20)}\)
\(=10.95 \mathrm{~m} / \mathrm{s}\)
Therefore, the speed of the car at \(t=10 \mathrm{~s}\) and \(t=20 \mathrm{~s}\) are \(24.5 \mathrm{~m} / \mathrm{s}\) and \(10.95 \mathrm{~m} / \mathrm{s}\) respectively.