How to derive the second equation of motion?
Asked by | 3rd Jun, 2008, 08:25: PM
Consider the definition of velocity. As we know, it is the rate of change of displacement i.e. velocity is the distance traveled per unit time. For ex: When we say that a car is traveling with a velocity of 80 km/hr what we actually mean is that the car is covering a distance of 80 km every hour. Note that here we assume that the velocity is uniform i.e. it remains 80 km/hr and doesn't change to any other value. Below is a car moving with a uniform velocity.
Thus. if x is the distance traveled by a body in a given time t, then the velocity v of the body is given by
Now we define acceleration as the rate of change of velocity i.e. the change in velocity itself per unit time. For example if a car is moving with a uniform velocity 80 km/hr and if in the next hour it's velocity becomes 100 km/hr, then we say that it has accelerated with an acceleration of 100-80 i.e. 20 km/hr. Note that the decrease in velocity is also an acceleration and is sometimes also called deceleration or retardation or negative acceleration. For ex: if the velocity of the car came down to 60 km/hr from 80 km/hr in one hour, then the acceleration of the car is said to be 60-80 i.e. -20 km/hr. The -ve sign indicates that the velocity of the car DECREASES by the given amount, every hour.
If u is the initial velocity of a moving body, and if the velocity of the body changes to v (which we shall call the final velocity) in a time interval t, then the acceleration of the body in the time t is given by
Note that in the above formula we assume that the acceleration of the body was uniform (i.e. the same) throughout the time interval t. Below is a car moving with a uniform acceleration.
Now, equation (0) can be re-written as
at = v-u
=> v-u = at
This is Newton's First equation of motion. As you can you see, we can use this equation to calculate the velocity of a body which underwent an acceleration of a m/s for a time period of t seconds, provided we know the initial velocity of the body. Initial velocity i.e. u is the velocity of the body just before the body started to accelerate i.e. the velocity at t=0.
In case, the body started to accelerate from rest then we can substitute the value of initial velocity to be u=0.
We sometimes also may want to find the total distance traveled by moving body.
A moving body might be either moving with a uniform velocity, or with a uniform acceleration or even with a non-uniform acceleration.
In case of a body moving with a uniform velocity v, it is quite simple to calculate the total distance s traveled by the body in a time t. we know that
velocity = distance traveled / time taken
v = s/t
=> s= vt
Thus, distance traveled = velocity x time
Now the situation is slightly different for a body moving with a uniform acceleration a. To calculate the distance traveled by an uniformly accelerating body, we derive the equation as follows.
If u is the initial velocity of an uniformly accelerating body and v is its velocity after a time t, then since the acceleration is UNIFORM, we can find the average velocity of the body as follows
average velocity = (u+v)/2
Now, the distance s, traveled in the time t by the body is given by
distance traveled = average velocity x time
s = [(u+v)/2]t
From equation (1) we have v=u+at, substituting this in the above equation for v, we get
s = [(u+u+at)/2]t
=> s = [(2u+at)/2]t
=> s = [(u + (1/2)at)]t
This is Newton's second equation of motion.
Answered by | 3rd Jun, 2008, 10:27: PM
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