Can anyone provide simplified derivations for equations of kinematics?

  • Thread starter Thread starter rohannet
  • Start date Start date
AI Thread Summary
Simplified derivations for the equations of kinematics are requested, specifically for v=u+at, s=ut+1/2at², v²-u²=2as, and Dn=u+a/2(2n-1). The original poster is in Class XI and struggles with calculus, seeking easier explanations. A suggestion is made to refer to textbooks for these derivations, and a link to HyperPhysics is provided for additional resources. The poster acknowledges the textbook content but expresses difficulty in understanding it. Clear and simplified explanations are essential for grasping these fundamental kinematic equations.
rohannet
Messages
8
Reaction score
0
I AM CURRENTLY STUDYING IN CLASS XI AND I DON'T KNOW MUCH ABOUT CALCULUS.
SO CAN ANYBODY GIVE EASY DERIVATIONS OF EQUATIONS OF KINEMATICS
1)v=u+at
2)s=ut+1/2at2
3)v2-u
PHP: 
2
=2as
4)Dn=u+a/2(2n-1)

THANKS IN ADVANCE
 
Physics news on Phys.org
thank u for Ur effort
i know that its all in my textbook
but that too i don't understand.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Two cars moving towards each other - bullets fired
Replies
6
Views
2K
What is the minimum velocity needed?
Replies
20
Views
2K
Proof of Kinematics Equation: Eliminating Time from the Equation
Replies
1
Views
1K
Using Energy and momentum conservation to derive the equation
Replies
3
Views
1K
Kinematic Equation Rearrangement
Replies
1
Views
3K
Having a problem in steps while solving integrals
Replies
6
Views
1K
Can You Solve the Problem of Projectiles with These Equations?
Replies
1
Views
1K
Correct Derivation of the Adiabatic Condition? (PV Diagram)
Replies
4
Views
995
Deriving Lorentz Transformation
Replies
4
Views
1K
Measurement technology textbook problems
Replies
1
Views
550

Hot Threads

Recent Insights

Back
Top