The force which helps to move in circle is known as centripetal force. Its a center seeking force and this force can only the direction of the car.
Centripetal acceleration = [tex]\frac{V^2}{R}[/tex]
[tex]a_c = 25.2 \frac{m}{s^2}[/tex]
R = 18.5 m
V = ?
Using the above formula:
[tex]V^2 = a_c \times R[/tex]
[tex]V^2 = 25.2 \times 18.5[/tex]
[tex]V^2 = 466.2[/tex]
[tex]V = \sqrt{466.2}[/tex]
V = 21.59 m/s
Hence, the speed of the car should be 21.59 m/s.
The speed of the car is about 21.6 m/s
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Centripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
[tex]\texttt{ }[/tex]
Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of car = m = 1000 kg
radius of the circular turn = R = 18.5 meters
centripetal acceleration = a = 25.2 m/s²
Unknown:
speed of the car = v = ?
Solution:
[tex]a = v^2 \div R[/tex]
[tex]v^2 = aR[/tex]
[tex]v = \sqrt{aR}[/tex]
[tex]v = \sqrt{25.2 \times 18.5}[/tex]
[tex]v \approx 21.6 \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
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Grade: High School
Subject: Physics
Chapter: Circular Motion
