Physics Help on Derivatives?

Question

In Physics, at least at high school level, we have a reference table with all the basic equations on it however in some problems, you need to derive from the basic formulas to get an answer. Those are the questions I usually have trouble with.


Can anyone explain to me in steps the types of derivatives, how tos, examples and etc. on basic derivation.

Oppps.... to be more precise:

i.e. A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at –7.0 meters per second2 to a full stop. Approximately how far does the roller coaster
travel during its deceleration?

The reference table gives us the equation:
vf^2= Vi^2+ 2ad

But we can't use it that way, we have to derive an equation to solve for d.

In the book it says,

d= vf^2-vi^2/2a

I would have never thought of using that. And there are more problems. It's dificult for me to rearrange the formula to get d. I don't think about deriving it and I get lost. Is there an easier way to explain it.

Answer 1

Good question. I have same problem. But no one can explane how to use derivative and Integration in physics. It is better to understand by yourself. For that you may read Solved Problems in highschool books. More you can refer popular books like; Concept by Halliday and Walker, Concept of physics ny Dr H C Verma. ( Indian author ) . Belive me these solved problems really program our brain for that kind of problems. Do not depend on any teacher. Learn yourself because you are a Excellent teacher for your self. THANK YOU.

Answer 2

derivative
is basically, taking the graph equation
and creating an equation which graphs the gradient of the original graph equation

the simple way to get the derivative
is to
take the power of x (or the variable) and multiply it to the front of the component, then -1 to the power, the same applies to each component of the equation ( there are 2 ways to write derivatives, either as dy/dx or f'(x))

where:
dy/dx x^6 = 6x^5
etc
so where an equation f(x) = 5x, f'(x) = 5
and where it is f(x) = 6, f'(x) = 0

should your equation come in 1/x^2
remember that 1/x^2 = x^-2
the same applies
making it -2x^-1 which = -2/x

so a sample question would be
find the derivative of the parabola with equation
f(x) = 5x^2 + 5x + 6
f'(x) = 10x + 5

you can graph this out, and see how the line corresponds with the gradient of the graph
In physics, the derivative is used for finding the acceleration from a velocity value

you may be asked to use first principles in maths, but not-so for physics

Answer 3

But you can use the formula like that. Take the formula
vf^2= Vi^2+ 2ad
and plug in the info you know
0^2 = (15 m/s)^2 + 2*(-7 m/s^2)*d
Then evaluate, using all the algebra you know
0 = 225 m^2/s^2 - (14 m/s^2)*d note 1
0 + (14 m/s^2)*d = 225 m^2/s^2 note 2
(14 m/s^2)*d = 225 m^2/s^2 note 3
d = (225 m^2/s^2) / (14 m/s^2) note 4
d = (225/16) (m^2/s^2)/(m/s^2) = 16 m
note 1: next move the 2nd term from right side to left
note 2: next drop the 0
note 3: next divide both sides by (14 m/s^2)
note 4: next simplify the magnitude and the units separately, cancell where you can.

Answer 4

acceleration is -7 m/s^2 means the velocity reduces at the rate of 7 per second. So in order to make the initial velocity 15 zero we have t produce the change of 15. Naturally this will b done in 15/7 seconds.. at t = 0 velocity was 15 m/s and after 15/7 seconds it was zero so the average velocity was [15+0]/2 = 7.5 m/s. So the distance travelled = average velocity multiplied by time. That is 7.5x(15/7) = 16.07 m. We can apply the last formula because the velocity is changing at a constant rate.

Answer 5

Any text book on first year university calculus should cover this, I recommend Stewart "Calculus" (4th ed +) it includes integral and differential calculus. The answers to the above are (unless I have made a mistake!): d/dx(300x^2)= 2*300x =600x d/dx(x^3+3)^2=d/dx((x^3+3)*(x^3+3))=d/... =6x^5 + 3*6x^2 +0 = 6x^5 + 18x^2 = 6x^2*(x^3+ 18) ie: d/dx (x^n)= n*x^n-1 and d/dx(c)= 0 where c is a constant I highly recommend taking the subject calculus as the other people in your class are, as it does get fairly indepth, and maths will give you a good background for physics.