
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to:
A. ${{t}^{\dfrac{3}{4}}}$
B. ${{t}^{\dfrac{3}{2}}}$
C. ${{t}^{\dfrac{1}{4}}}$
D. ${{t}^{\dfrac{1}{2}}}$
Answer
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Hint: You could begin from the definition of power and then substitute the expression for work done in it. Now, you could use Newton's law to substitute for the force. With the help of definition of acceleration, you could express the power in terms of displacement. Doing so, you will end up with an equation with all the other terms except the displacement and time are constant and now you could find the required proportionality relation.
Formula used:
Expression for power,
$P=\dfrac{W}{t}$
Expression for work done,
$W=F\times S$
Newton’s second law,
$F=ma$
Complete step by step answer:
We are given a body moving along a straight line by a machine which is delivering constant power.
So, we should note that the power given here is constant.
Now let us define power. It is the time rate at which we do a work or the time rate at which we transfer energy. That is,
$P=\dfrac{W}{t}$ …………….. (1)
But we know that the work done is given by the product of force and displacement. That is,
$W=F\times S$ ……………. (2)
Where, F is the force and S is the resultant displacement.
Substituting (2) in (1) we get,
$P=\dfrac{F\times S}{t}$ ………………. (3)
Now, from Newton’s second law of motion, we have,
$F=ma$ …………. (4)
Substituting (4) in (3) we get,
$P=\dfrac{\left( ma \right)\times S}{t}$ ……………. (5)
We know that acceleration is the time rate of change of velocity. So,
$a=\dfrac{v}{t}$
Velocity is again the time rate of change of displacement, that is,
$v=\dfrac{S}{t}$
So, acceleration now becomes,
$a=\dfrac{S}{{{t}^{2}}}$ ……………………… (6)
Substituting (6) in (5) we get,
$P=\dfrac{m\times \left( \dfrac{S}{{{t}^{2}}} \right)\times S}{t}$
$P=\dfrac{m{{S}^{2}}}{{{t}^{3}}}$ ………………….. (7)
We are to find the proportionality relation of the distance moved by the body (S) with the time taken (t). Rearranging equation (7), we get,
${{S}^{2}}=\dfrac{P{{t}^{3}}}{m}$ ……………… (8)
We are given in the question that constant power is delivered, which implies the power here is constant. Also, we know that, mass of the body remains constant throughout the motion. So, we get P and m as constant in equation (8). We could now express (8) as,
${{S}^{2}}\propto {{t}^{3}}$
That is,
$S\propto {{t}^{\dfrac{3}{2}}}$
Hence, the solution to the given question is option B.
Note:
If you knew that that the product of force and velocity gives the power of a body, you could begin deriving from there. Doing so will save you time. While finding the proportionality relation between two terms in an equation, before converting the equation into proportionality, make sure that all other terms are constant.
Formula used:
Expression for power,
$P=\dfrac{W}{t}$
Expression for work done,
$W=F\times S$
Newton’s second law,
$F=ma$
Complete step by step answer:
We are given a body moving along a straight line by a machine which is delivering constant power.
So, we should note that the power given here is constant.
Now let us define power. It is the time rate at which we do a work or the time rate at which we transfer energy. That is,
$P=\dfrac{W}{t}$ …………….. (1)
But we know that the work done is given by the product of force and displacement. That is,
$W=F\times S$ ……………. (2)
Where, F is the force and S is the resultant displacement.
Substituting (2) in (1) we get,
$P=\dfrac{F\times S}{t}$ ………………. (3)
Now, from Newton’s second law of motion, we have,
$F=ma$ …………. (4)
Substituting (4) in (3) we get,
$P=\dfrac{\left( ma \right)\times S}{t}$ ……………. (5)
We know that acceleration is the time rate of change of velocity. So,
$a=\dfrac{v}{t}$
Velocity is again the time rate of change of displacement, that is,
$v=\dfrac{S}{t}$
So, acceleration now becomes,
$a=\dfrac{S}{{{t}^{2}}}$ ……………………… (6)
Substituting (6) in (5) we get,
$P=\dfrac{m\times \left( \dfrac{S}{{{t}^{2}}} \right)\times S}{t}$
$P=\dfrac{m{{S}^{2}}}{{{t}^{3}}}$ ………………….. (7)
We are to find the proportionality relation of the distance moved by the body (S) with the time taken (t). Rearranging equation (7), we get,
${{S}^{2}}=\dfrac{P{{t}^{3}}}{m}$ ……………… (8)
We are given in the question that constant power is delivered, which implies the power here is constant. Also, we know that, mass of the body remains constant throughout the motion. So, we get P and m as constant in equation (8). We could now express (8) as,
${{S}^{2}}\propto {{t}^{3}}$
That is,
$S\propto {{t}^{\dfrac{3}{2}}}$
Hence, the solution to the given question is option B.
Note:
If you knew that that the product of force and velocity gives the power of a body, you could begin deriving from there. Doing so will save you time. While finding the proportionality relation between two terms in an equation, before converting the equation into proportionality, make sure that all other terms are constant.
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