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For a spinning amusement park ride, the velocity v in meters per second of a car moving around a curve with a radius r meters is given by the formula v=sqrt(ar), where 'a' is the car’s acceleration in m/s^2.

AFor safety reasons, a ride has a minimum acceleration of 39.2m/s^2 . If the cars on the ride have a velocity of 14m/s what is the smallest radius that any curve on the ride may have?

BWhat is the acceleration of a car moving at 8m/s around a curve with a radius of 2.5m?

Verified Answer :

a)
v=sqrt(ar)
v^2=ar
r=(v^2)/a
r=(14^2)/39.2
r=5

5 m

b)
v=sqrt(ar)
v^2=ar
a=(v^2)/r
a=(8^2)/2.5
a=25.6

25.6ms^-2

shubhamchouhanvrVT

The smallest radius will be, r = 5 m. And the acceleration of a car is, a = 25.6 m/s².

What is acceleration?

Acceleration is the ratio of the change of the velocity with respect to time or the rate of change of the velocity.

a.The relation between the velocity, the radius of the path, and the acceleration on the circular path is given by :

Acceleration, a = 39.2 m/s²

Velocity of the car, v = 14 m/s

The radius of the path can be calculated using the above formula i.e.

\(r=\dfrac{v^2}{a}\)

\(r=\dfrac{(14)^2}{39.2}\)

r = 5 m

b. Velocity of the car, v = 8 m/s

The radius of the path, r = 2.5 m

The acceleration of the car is given by :

\(a=\dfrac{8^2}{2.5}\)

a = 25.6 square meter.

Therefore, the smallest radius will be, r = 5 m. And the acceleration of a car is, a = 25.6 m/s².

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