Question

VP 3.12.1 Part APart complete A cyclist going around a circular track at 10.0 m/s has a centripetal acceleration of 5.00 m/s2. What is the radius of the curve? Express your answer with the appropriate units. R = 20.0 m Previous Answers Correct VP 3.12.2 Part B A race car is moving at 40.0 m/s around a circular racetrack of radius 265 m. Calculate the period of the motion. Express your answer in seconds. T = nothing s Request Answer Part C Calculate the car’s centripetal acceleration.

Answer 1

Answer:

A) r = 20.0 m

B) T = 41.6 s

C) = 6.1 m/s²

Explanation:

A)

• The centripetal acceleration is the one that explains that even though the cyclist is moving at a constant speed, his velocity is changing the direction all the time, keeping him around a circle.
• This acceleration can be expressed as follows:

        $a_{c} =\frac{v^{2}}{r} = \frac{(10.0m/s)^{2}}{r} = 5.00 m/s2 (1)$

• Solving for r:

       $r = \frac{v^{2}}{a_{c} } = \frac{(10.0m/s)^{2}}{5.00m/s2} = 20.0 m (2)$

B)

• We can apply the definition of linear velocity, remembering that the period is the time needed to complete an entire circle (T).
• The arc around a circumference (the distance traveled) , is just 2*π*r, so applying the definition of linear velocity, we can write the following expression:

        $v = \frac{\Delta s}{\Delta t} = \frac{2*\pi*r}{T} (3)$

• Solving  for T:

       $T = \frac{\Delta s}{v} = \frac{2*\pi*r}{v} = \frac{2*\pi*265m}{40.0m/s} =41.6 s (4)$

C)

• The centripetal acceleration of the car from B) can be found as follows:

        $a_{c} =\frac{v^{2}}{r} = \frac{(40.0m/s)^{2}}{265m} = 6.1 m/s2 (5)$