
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
528.6k+ views
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.
Recently Updated Pages
Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 6 Maths: Engaging Questions & Answers for Success

Class 6 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

