The relation between time (T) and distance (X) $t = a{x^2} + bx$ where a and b are constants. Express instantaneous acceleration in terms of instantaneous velocity.
Answer
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Hint: Recall that velocity is defined as the rate of change of displacement of a particle with respect to time. It is a vector quantity. So it has both magnitude and direction. Also acceleration is the rate of change of velocity of a particle with time. It is also a vector quantity.
Complete answer:
Step I:
The given equation of time and distance relation is
$ \Rightarrow t = a{x^2} + bx$---(i)
Where ‘a’ and ‘b’ are constants
‘x’ is the distance and ‘t’ is the time.
Differentiating equation (i) with respect to time,
\[ \Rightarrow 1 = 2ax\dfrac{{dx}}{{dt}} + b\dfrac{{dx}}{{dt}}\]
\[ \Rightarrow 1 = (2ax + b)\dfrac{{dx}}{{dt}}\]
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{2ax + b}}$---(ii)
Step II:
Therefore, the velocity of the particle can be written as
$ \Rightarrow v = \dfrac{{dx}}{{dt}}$---(iii)
Comparing (ii) and (iii), the equation can be written as
$ \Rightarrow v = \dfrac{1}{{2ax + b}}$---(iv)
Step III:
Similarly, acceleration of the particle can be written as
$ \Rightarrow a = \dfrac{{dv}}{{dt}}$
Substituting the value of ‘v’, from equation (iv), and differentiating,
$ \Rightarrow a = \dfrac{d}{{dt}}(\dfrac{1}{{2ax + b}})$
$ \Rightarrow a = \dfrac{{ - (2a\dfrac{{dx}}{{dt}})}}{{{{(2ax + b)}^2}}}$
From equation (iii), substitute the value in the above equation, and solving
$ \Rightarrow a = \dfrac{{ - 2av}}{{{{(2ax + b)}^2}}}$
Again from equation (iv) it is clear that, ${v^2} = \dfrac{1}{{{{(2ax + b)}^2}}}$
Therefore, acceleration of the particle is $a = - 2a{v^3}$
Step IV:
Hence, the instantaneous acceleration of the particle in terms of instantaneous velocity is $ - 2a{v^3}$
Note:
It is to be remembered that instantaneous velocity of a particle is defined as the velocity of the particle at a particular instant of time. It can be positive, negative or zero. The instantaneous speed of an object is equal to instantaneous velocity. When any object is changing its acceleration, at different intervals of time, then the acceleration of the object at a particular instant of time is known as instantaneous acceleration.
Complete answer:
Step I:
The given equation of time and distance relation is
$ \Rightarrow t = a{x^2} + bx$---(i)
Where ‘a’ and ‘b’ are constants
‘x’ is the distance and ‘t’ is the time.
Differentiating equation (i) with respect to time,
\[ \Rightarrow 1 = 2ax\dfrac{{dx}}{{dt}} + b\dfrac{{dx}}{{dt}}\]
\[ \Rightarrow 1 = (2ax + b)\dfrac{{dx}}{{dt}}\]
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{2ax + b}}$---(ii)
Step II:
Therefore, the velocity of the particle can be written as
$ \Rightarrow v = \dfrac{{dx}}{{dt}}$---(iii)
Comparing (ii) and (iii), the equation can be written as
$ \Rightarrow v = \dfrac{1}{{2ax + b}}$---(iv)
Step III:
Similarly, acceleration of the particle can be written as
$ \Rightarrow a = \dfrac{{dv}}{{dt}}$
Substituting the value of ‘v’, from equation (iv), and differentiating,
$ \Rightarrow a = \dfrac{d}{{dt}}(\dfrac{1}{{2ax + b}})$
$ \Rightarrow a = \dfrac{{ - (2a\dfrac{{dx}}{{dt}})}}{{{{(2ax + b)}^2}}}$
From equation (iii), substitute the value in the above equation, and solving
$ \Rightarrow a = \dfrac{{ - 2av}}{{{{(2ax + b)}^2}}}$
Again from equation (iv) it is clear that, ${v^2} = \dfrac{1}{{{{(2ax + b)}^2}}}$
Therefore, acceleration of the particle is $a = - 2a{v^3}$
Step IV:
Hence, the instantaneous acceleration of the particle in terms of instantaneous velocity is $ - 2a{v^3}$
Note:
It is to be remembered that instantaneous velocity of a particle is defined as the velocity of the particle at a particular instant of time. It can be positive, negative or zero. The instantaneous speed of an object is equal to instantaneous velocity. When any object is changing its acceleration, at different intervals of time, then the acceleration of the object at a particular instant of time is known as instantaneous acceleration.
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