For a spinning amusement park ride, the velocity [tex]v[/tex] in meters per second of a car moving around a curve with a radius [tex]r[/tex] meters is given by the formula [tex]v = \sqrt{ar}[/tex], where [tex]a[/tex] is the car’s acceleration in [tex]m/s^2[/tex].
a) For safety reasons, a ride has a minimum acceleration of [tex]39.2 \, m/s^2[/tex]. If the cars on the ride have a velocity of [tex]14 \, m/s[/tex], what is the smallest radius that any curve on the ride may have?
b) What is the acceleration of a car moving at [tex]8 \, m/s[/tex] around a curve with a radius of [tex]2.5 \, m[/tex]?
Answer (3)
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 = a v 2
r = 39.2 ( 14 ) 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 = 2.5 8 2
a = **25.6 **square meter.
Therefore, the smallest radius will be , r = 5 m. And t he acceleration of a car is, a = 25.6 m/s².
To know more about **acceleration **follow
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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
The smallest radius that any curve on the ride may have is approximately 5.0 meters. The acceleration of a car moving at 8 m/s around a curve with a radius of 2.5 meters is 25.6 m/s².
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