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**1. Homework Statement**

Hi, Im taking a correspondance physics course and just wanted to know if im on the right track with understanding Newton's principles.

Roberta drives her 1452 kg car along a straight , level road at a consant velocity of 30.0m/s (E). Her brakes suddenly give out. She puts the car in neutral and lets it coast for 25.0 s. The air drag deccelerates the car to a velocity of 25.0 m/s (E). (Assume a frictionless surface. )

(a) Determine the average acceleration while the car is decelerating.

(b) Determine the average force of the air against the car.

(c) After coasting for 25.0 s , Roberta pulls her handbrake to slow the car to a stop. If it takes 3.0 s to stop the car, what is the force applied by the handbrake. (Assume that the force exerted by the air remains constant and is equal to the force determined in part (c).

**2. Homework Equations**

F=ma

F(v) = mg

d=1/2 (v1 + v2)(t)

a = F app/ m

**3. The Attempt at a Solution**

(a) Given

m=1452 kg

force of air resistance = 25.0 m/s E

Required: acceleration

Analysis: F=m(a)

F(net) = F(app) = ma

a = F app/ m

=25.0 m/s/1452 kg

=0.0172 m/s E

(b) Given:

m=1452 kg

v1 = 30.0 m/s E

v2 = 25.0 m/s E

need to also find the distance:

d=1/2 (v1 + v2) (t)

=1/2 (30 m/s + 25 m/s) (25)

687.5 m

Required: F(air)

Analysis: v2^2 = v1^2 + 2a(d)

a= v2^2-v1^2 / 2(d)

=(25m/s)^2 - (30.0 m/s)^2 / 2(687.5 m

=-0.2 m/s^2

F air = m(a)

=(1452kg)(-0.2 m/s^2)

=-290. 4 N

(c) Given = m=1452 kg

g= 25.0 m/s E

Required F(n)

Analysis: Fnet = Fn-mg

Fv = mg

Fv = (1452kg) (25.0m/s)

=36300 N

(d) d= 1/2(V1 + V2) (t)

1/2(30 m/s - 25 m/s)(3)

=7.5 m(after she applies the handbrake )

d= 1/2(30-25 m/s) (25)

=62.5 m (distance covered coasting for 25 s)

therefore I would add the two distances together and the total displacment from the time her brakes give out to the time she stops = 70 m .

Any feedback would be apprecaited

Thanks

Pharm 89