Question

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's acceleration?

2.0 m/s/s
2.5 m/s/s
5.0 m/s/s
20. m/s/s
50. m/s/s

Answer 1

I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s

Answer 2

a = 20 m/s²

Step-by-step explanation:

The magnitude of the acceleration of an object that moves in a circular path is given by:

`a= \sqrt{(a_(t))^(2) +(a_(n))^(2) }` Formula (1)

where:

at = α*R : Formula (2) :Tangential acceleration

ac = ω²*R : Formula (3) : Centripetal acceleration

α : angular acceleration (rad/s²)

ω: angular speed (rad/s)

R : is radius where the object is located from the center of the circular path

The tangential velocity of the body is calculated as follows:

v = ω*R Formula (4)

where:

v is the tangential velocity or linear velocity (m /s)

ω is the angular speed (rad/s)

R is radius where the body is located from the center of the circular path

Data

v = 10 m/s : tangential speed of the object (uniform)

R = 10 m

Calculating of ω (angular speed)

We replace data in the formula (4)

v = ω*R

10 = ω*5

ω = 10 / 5

ω = 2 rad/s

Calculating of the Centripetal acceleration (ac)

We replace ω = 2 rad/s in the formula (3)

ac = ω²*R

ac = (2)²*(5)

ac = 20 m/s²

Calculating of the tangential acceleration (at)

Because the magnitude of the tangential speed is uniform , then α=0

We replace α=0 in the formula (2)

at =α*R = 0*5

at = 0

Calculating of the magnitude of the object's acceleration (a)

We replace : at = 0 and ac = 20 m/s² in the formula (1)

`a= \sqrt{(0)^(2) +(20)^(2) }`

a = 20 m/s²