Need help with Physics homework

[max]

Physics problems involving calculus

Can anyone point me in the right direction as to how I should try to solve these problems? I show my attempts at solving them but I am not sure if I am solving them correctly. Also, I'm not sure if my use of significant digits are correct. Any help would be appreciated.

Over a period of (20+-1)s, a car changes from moving forward at (12+-1)m/s to moving (8+-)m/s in reverse. Find the average acceleration of this car.

my answer (8+-1 - 12+-1)/(20+-1)s = (-4+-2)/(20+-1) = (-0.2+-0.1)m/s^2

Consider an object with position given by s=2.27t^3-1.81t+1.12, where position is in m when time is in s. a)Find its average velocity from t=0.00 s to t=4.70 s. b)Find its velocity at t=2.35 s. c)Find the time at which the object velocity is zero. d)Find the objects acceleration at t=2.35 s.

my answers
s=2.27t^3-1.81t+1.12
v=6.81t^2-1.81
a=13.62t

a) (96.45-(1.21))/4.70 = 20.3m/s I believe average velocity is displacement over change in time.
b) Just plugging in t=2.35 in velocity equation to get 35.8m/s
c) 6.81t^2-1.81=0, t= +-0.516, we want positive value for time can't be negative.
d) Just plugging in t=2.35 in acceleration equation to get 32.0

A moving object has a velocity v=1.2t-0.23t^2, where the velocity is in m/s when time is in seconds. a)Find the object's maximum velocity. b)Given that the object is at the origin at t=0, determine its location when the velocity is a maximum. c)Find the object's acceleration when its velocity is zero.

my answers
v=1.2t-0.23t^2
Integrating to get s=0.6t^2-0.077t^3, where C=0
a=1.2-0.46t

a) I think max velocity occurs when acceleration = 0. This happens at t=1.2/.46= 2.6. Then I plug this time into velocity equation to get v=1.6
b) Since I just found the maximum velocity, I would plug that into the position equation to get 1.220608 or 1m
c) I set velocity equation= 0 to get t(1.2-0.23t)=0, t=0 and t=5.2. Plugging those numbers into the acceleration equation yields a=1.2 and a= -1.2

 

[DrPhil]

Though not stated as required in the problem, I tried to round numbers in calculation to the correct number of significant figures.

Over a period of (20+-1)s, a car changes from moving forward at (12+-1)m/s to moving (8+-)m/s in reverse. Find the average acceleration of this car.

answer (8+-1 - 12+-1)/(20+-1)s = (-4+-2)/(20+-1) = (-0.2+-0.1)m/s^2

What are they teaching you about propagation of errors!!? If the uncertainties are independent and random, then when you add or subtract, the errors should add quadratically (square root of sum of squares). [If they haven't taught you this, they should! As a professional physicist, correctly propagating statistical errors is one of my pet peeves.]

More important, you failed to note that the final speed is "in reverse," which makes it negative.
\(\displaystyle [- (8 \pm 1)m/s - (12 \pm 1)m/s] = (- 20 \pm 1.4)m/s\)

When you multiply or divide, the relative (percentage) errors add in quadrature:
\(\displaystyle \dfrac {- (20 \pm 1.4)m/s}{(20 \pm 1)s} = - (1.0 \pm 0.09)m/s^2\)

Hmmm. Your final error came out correct anyhow.

I don't have time to check the other two right now, but will return later. [I don't agree with the velocity result in part (a), because I have a different value for s(4.70s).]

 

[max]

What are they teaching you about propagation of errors!!? If the uncertainties are independent and random, then when you add or subtract, the errors should add quadratically (square root of sum of squares). [If they haven't taught you this, they should! As a professional physicist, correctly propagating statistical errors is one of my pet peeves.]

More important, you failed to note that the final speed is "in reverse," which makes it negative.
\(\displaystyle [- (8 \pm 1)m/s - (12 \pm 1)m/s] = (- 20 \pm 1.4)m/s\)

When you multiply or divide, the relative (percentage) errors add in quadrature:
\(\displaystyle \dfrac {- (20 \pm 1.4)m/s}{(20 \pm 1)s} = - (1.0 \pm 0.09)m/s^2\)

Hmmm. Your final error came out correct anyhow.

I don't have time to check the other two right now, but will return later. [I don't agree with the velocity result in part (a), because I have a different value for s(4.70s).]

Thank you so much with the reply and thanks for being willing to review the other problems.

For the propagation of error, I did it the way it shows it in my physics text, physics for scientists and engineers 9th edition on page A21 in the back of the book.

Your way seems more precise, I wonder why my text does it differently.

 

[HallsofIvy]

Can anyone point me in the right direction as to how I should try to solve these problems? I show my attempts at solving them but I am not sure if I am solving them correctly. Also, I'm not sure if my use of significant digits are correct. Any help would be appreciated.

Over a period of (20+-1)s, a car changes from moving forward at (12+-1)m/s to moving (8+-)m/s in reverse. Find the average acceleration of this car.

my answer (8+-1 - 12+-1)/(20+-1)s = (-4+-2)/(20+-1) = (-0.2+-0.1)m/s^2

Not quite, you need to be more careful with the "+-" part. The maximum forward speed is 12+1= 13 mph. The maximum reverse speed is -8-1= -9 mph. The largest possible change in speed is (-9)- 13= -22. The lowest forward speed is 12-1= 11 mph. The lowest reverse speed is -8+1= -7 mph. The least possible change in speed is -7- 11= -18.

The largest change in time during which this change in speed is made is 20+ 1= 21 seconds. The smallest change in time during which this speed in time during which this change in speed is made is 20- 1= 19 seconds. Dividing the largest change in speed by the smallest time change gives the greatest acceleration, -22/19 mile per hour per second. Dividing the smallest change in speed by the largest time change gives the least acceleration, -18/21 miles per hour per second. The average acceleration is (-22/19- (-18/21))/2.

Consider an object with position given by s=2.27t^3-1.81t+1.12, where position is in m when time is in s. a)Find its average velocity from t=0.00 s to t=4.70 s. b)Find its velocity at t=2.35 s. c)Find the time at which the object velocity is zero. d)Find the objects acceleration at t=2.35 s.

my answers
s=2.27t^3-1.81t+1.12
v=6.81t^2-1.81
a=13.62t

Yes, those are correct- although one could argue that, looking at signifiicant figures, the last should be a= 13.6t.

a) (96.45-(1.21))/4.70 = 20.3m/s I believe average velocity is displacement over change in time.

Your "belief" is correct but where did you get "96.45"? That is not what I get putting t= 4.7 into the equation for s.

b) Just plugging in t=2.35 in velocity equation to get 35.8m/s

Yes, that is correct.

c) 6.81t^2-1.81=0, t= +-0.516, we want positive value for time can't be negative.

Yes, that is correct.

d) Just plugging in t=2.35 in acceleration equation to get 32.0

Yes, that is correct- and the correct number of significant figures!

A moving object has a velocity v=1.2t-0.23t^2, where the velocity is in m/s when time is in seconds. a)Find the object's maximum velocity. b)Given that the object is at the origin at t=0, determine its location when the velocity is a maximum. c)Find the object's acceleration when its velocity is zero.

my answers
v=1.2t-0.23t^2
Integrating to get s=0.6t^2-0.077t^3, where C=0
a=1.2-0.46t

a) I think max velocity occurs when acceleration = 0. This happens at t=1.2/.46= 2.6. Then I plug this time into velocity equation to get v=1.6

Yes, that is correct.

b) Since I just found the maximum velocity, I would plug that into the position equation to get 1.220608 or 1m

Your position equation is correct but you put in t, not v!! Set t= 2.6, not 1.6.

c) I set velocity equation= 0 to get t(1.2-0.23t)=0, t=0 and t=5.2. Plugging those numbers into the acceleration equation yields a=1.2 and a= -1.2

Yes, that is correct.

 

[DrPhil]

For the propagation of error, I did it the way it shows it in my physics text, physics for scientists and engineers 9th edition on page A21 in the back of the book.

Your way seems more precise, I wonder why my text does it differently.

HallsofIvy showed how to manipulate errors with the definition of \(\displaystyle \pm \ \delta\) being that no random fluctuation of the measurement ever exceeds \(\displaystyle \delta\). In scientific practice, on the other hand, unless some other criterion is stated explicitly, the value used for the statistical uncertainty is \(\displaystyle \pm \ \sigma\), where \(\displaystyle \sigma\) is the standard deviation assigned to the precision of the measurement. That is what leads to the rules I showed.