Newton's Laws of Motion on a car

[pharm89]

Homework Statement

Hi, I am taking a correspondance physics course and just wanted to know if I am 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 let's 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).

Homework Equations

F=ma
F(v) = mg
d=1/2 (v1 + v2)(t)
a = F app/ m

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 displacement from the time her brakes give out to the time she stops = 70 m .
Any feedback would be apprecaited
Thanks
Pharm 89

 

[pharm89]

Homework Statement

Hi, I am taking a correspondance physics course and just wanted to know if I am 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 let's 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).

Homework Equations

F=ma
F(v) = mg
d=1/2 (v1 + v2)(t)
a = F app/ m

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 displacement from the time her brakes give out to the time she stops = 70 m .
Any feedback would be apprecaited
Thanks
Pharm 89

ps i forgot to add the last question in: (d) What is Roberta's displacement from the time her brakes give out to the time she stops?
Thanks

 

[Dorothy Weglend]

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

Does the (E) stand for East? I guess?

You are mistakenly using the velocity of the car at the end of the acceleration as the force of the air drag. You can see this right away if you focus on the units. Force has units of Newtons, while velocity has units of distance/time (m/s).

The simplest way to calculate the average acceleration is (change of v)/(amount of time). If you do it properly, it will be a negative number.

I don't think you need the second law here at all.

(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

This is right. Good job! But a little complicated. If you calculate the correct average acceleration in (a), you can just use this directly with F = ma, as you did at the end. Your calculation of the average acceleration isn't wrong, it's just a very long path to your destination.

(

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

Here you seem to be confusing velocity with acceleration. Velocity has units of distance/time (m/s), and acceleration has units of distance/time/time or distance/time^2 (m/s^2). I would also suggest a more careful reading of the problem.

After she applies the handbrake, the car stops in 3.0 s. The net force that slows the car is F_Net = F_brake + F_air. You can calculate F_net (average) by finding the average acceleration, just as you did in part a, and continue as in part b to find F_Net. Then you should be able to get F_brake from this last relationship.

(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 displacement from the time her brakes give out to the time she stops = 70 m .

If you have a negative velocity, it means that the car has actually reversed direction and is going backwards. That's not what is happening here, the car keeps traveling in the same direction.

In any case, the first interval is the coasting portion, which takes 25 seconds, and the initial and final velocities are 30 m/s and 25 m/s (not 30 m/s and -25 m/s).

The second interval is the handbrake, which takes 3 s, and the initial and final velocities are 3 m/s and 0 m/s.

Hope this was some help.

Dorothy

 

[pharm89]

Thanks very much for the help Dorothy. I hope now I have come up with the proper solutions.
(a) average acceleration is -0.2 m/s^2
(c) The car goes from 25 m/s to 0 m/s in 3 sec. Therefore change in velocity is 25 m/s and change in time is 3 sec. Therefore there is a deceleration of 25/3 = -8.33 m/s^2
Then i plugged this info into the equation:
F=m(a)
290.4 N + force of F(handbrake) = 1452 kg X 8.33 m/s^2
290.4 N + F(handbrake) = 12095.2
Therefore, F(Handbrake) = 12095.2-290.4 = 1180.8 N

(d)1st interval is the coasting portion:
d= 1/2(v1 + v2) (t)
=1/2 (30 +25) (25)
=1/2 (55)(25)
=687.5 m

2nd interval is the handbrake:
d= 1/2 (v1 + v2) (t)
=1/2 (3 m/s + 0m/s) (3)
=4.5 m

Therefore total displacement would be 692 m...I hope!
Thanks
Pharm 89

 

[Dorothy Weglend]

I didn't check your arithmetic, but that all looks good to me.

Good physics for 2007 Pharm 89 :-)

Dorothy

 

[Dorothy Weglend]

Oh Whoops. My mistake, Pharm89. Sorry. The velocities for the handbrake interval are 25 m/s and 0 m/s. Whoops :-(

I still hope you have a good physics year in 2007 :-)

Dorothy