The velocity of an object traveling in a circle is quadrupled and its radius is tripled The acceleration of this object will change by factor of?

The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.

Sample Problem #1

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 900 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

a =

Fnet =

To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (10.0 m/s)2 / (25.0 m)

a = (100 m2/s2) / (25.0 m)

a = 4 m/s2

To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.

Fnet = m • a

Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

Sample Problem #2

A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 95.0 kg

R = 12.0 m

Traveled 1/4-th of the circumference in 2.1 s

Requested Information:

v =

a =

Fnet =

To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:

v = d / t

v = (0.25 • 2 • pi • R) / t

v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)

v = 8.97 m/s

To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (8.97 m/s)2 / (12.0 m)

a = (80.5 m2/s2) / (12.0 m)

a = 6.71 m/s2

To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.

Fnet = m*a

Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N

In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.

The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.

Sample Problem #1

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 900 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

a =

Fnet =

To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (10.0 m/s)2 / (25.0 m)

a = (100 m2/s2) / (25.0 m)

a = 4 m/s2

To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.

Fnet = m • a

Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

Sample Problem #2

A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 95.0 kg

R = 12.0 m

Traveled 1/4-th of the circumference in 2.1 s

Requested Information:

v =

a =

Fnet =

To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:

v = d / t

v = (0.25 • 2 • pi • R) / t

v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)

v = 8.97 m/s

To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (8.97 m/s)2 / (12.0 m)

a = (80.5 m2/s2) / (12.0 m)

a = 6.71 m/s2

To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.

Fnet = m*a

Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N

In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.

answer from:

mahmoud3321

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6