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In summary, the conversation discusses different methods for calculating the distance traveled by a body at contact acceleration. The first method uses the definition of acceleration and the second method uses the average velocity during each time period. The conversation also mentions solving a second order ODE as a more straightforward method.
- #1
vijay_singh
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Hi
I tried to derive the distance traveled by a body at contact acceleration from the definition of acceleration (increase in speed every sec), but the ended with a different result. Can you see what I am doing wrong.
u = initial speed
t = time taken
S = {distance in 1st sec} + {distance in 2nd sec } + {distance in 3rd sec) + ... + {distance in t sec}
S = {u } + {u + a} + {u + 2a } + ...... + {u + (t - 1)a}
S = u * t + {a + 2a + ...+ (t - 1)a }
S = ut + a( 1 + 2 + 3 ...+ (t-1))
S = ut + a * t * (t -1) / 2
Vijay
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- #2
rcgldr
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vijay_singh said:
S = {u } + {u + a} + {u + 2a } + ...... + {u + (t - 1)a}
The distance moved is based on average velocity during each period, not the final velocity at the end of each time period:
S = {u + (1/2)a } + {u + (3/2)a} + {u + (5/2)a } + ... + {u + ((2t-1)/2)a}
To calculate the sum of the coefficients for a:
Code:
c = ( 1)/2 + ( 3)/2 + ( 5)/2 + ... + (2t-5)/2 + (2t-3)/2 + (2t-1)/2 2c = ( 1)/2 + ( 3)/2 + ( 5)/2 + ... + (2t-5)/2 + (2t-3)/2 + (2t-1)/2 + (2t-1)/2 + (2t-3)/2 + (2t-5)/2 + ... + ( 5)/2 + ( 3)/2 + ( 1)/2 ------------------------------------------------- (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 = t^2c = 1/2 t^2S = u * t + 1/2 * a * t^2
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- #3
Hootenanny
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A much more straight forward method would be to solve the second order ODE:
[tex]\frac{d^2s}{dt^2} = \text{const.}[/tex]
Related to Derivation of S = ut + 1/2at^2
1. What is the meaning of the equation S = ut + 1/2at^2?
The equation S = ut + 1/2at^2 is used to calculate the displacement (S) of an object in motion with a constant acceleration (a). It takes into account the initial velocity (u) and the time (t) that the object has been in motion.
2. Why is the equation called "Derivation of S = ut + 1/2at^2"?
The equation is called "derivation" because it is derived from the equations of motion, specifically the equation for displacement (S = ut + 1/2at^2). It is a result of combining the equations for velocity (v = u + at) and acceleration (a = (v-u)/t).
3. What are the units for the variables in the equation S = ut + 1/2at^2?
The units for displacement (S) are usually in meters (m). The units for initial velocity (u) and final velocity (v) are in meters per second (m/s). The unit for time (t) is in seconds (s). The unit for acceleration (a) is usually in meters per second squared (m/s^2).
4. Can the equation S = ut + 1/2at^2 be used for any type of motion?
Yes, the equation can be used for any type of motion, as long as the acceleration is constant. This includes both linear and projectile motion.
5. How can the equation S = ut + 1/2at^2 be used in real-life applications?
The equation can be used in various fields such as physics, engineering, and sports. For example, it can be used to calculate the distance traveled by a car in a given time, the height reached by a projectile, or the trajectory of a ball in a sport like basketball or soccer.
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