
A machine is delivering constant power to drive a body along a straight line. What is the relation between the distance traveled \[\left( S \right)\] by the body against time \[\left( t \right)\] ?
A. \[{S^2} \propto {t^3}\]
B. \[{S^2} \propto {t^{ - 3}}\]
C. \[{S^2} \propto {t^2}\]
D. \[S \propto {t^3}\]
Answer
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Hint: Work is a force exerted across a distance in physics, and power is the pace at which work is done. In other words, the product of force and velocity is known as Power. Here power is constant, so we need a relation connecting distance and time in a constant power scenario. Try to write power in terms of distance and time.
Complete answer:
Here we have asked to find a relation between distance \[\left( S \right)\] and time \[\left( t \right)\] in constant Power \[P\] .
The relation between Power, Force, and Velocity is,
\[P = F \times V\] …………..(1) Where \[P = \] Power; \[F = \] Force; \[V = \] Velocity
Let us divide the equation (1) into two-part as, one of the Force part and the other of the Velocity part. Coming to the Force part, let us write the equation of force as,
\[F = m \times a\]………(2), where \[m = \] mass and \[a = \] acceleration. And acceleration can be written as,
\[a = \dfrac{V}{t}\]…….(3) and \[V = \dfrac{S}{t}\]…….(4), where \[V = \]Velocity; \[S = \] distance, and \[t = \] time.
Comparing equations (3) and (4) it can be written as,
\[a = \dfrac{S}{{{t^2}}}\]………..(5). Substituting (5) in (2) we get,
\[F = m \times \dfrac{S}{{{t^2}}}\]………(6). Rewriting equation (1) in terms of (4) and (6),
\[P = m \times \dfrac{S}{{{t^2}}} \times \dfrac{S}{t}\]
\[ \Rightarrow Pm{t^3} = {S^2}\].
As Power and mass are constants,
\[ \Rightarrow {S^2} \propto {t^3}\]
That square of distance is proportional to the cube of the distance at constant Power.
Hence option (A) is correct.
Note: Power is a scalar quantity. That is, it has no direction but the only magnitude. Hence it is the scalar product of Force and Velocity.
At constant Power, \[V = \sqrt t \] where \[V = \] velocity and \[t = \]time. ( Can be derived from above equations)
Unit of Power is Watt(W). But horsepower unit is a standard measure for describing the power capabilities of daily equipment, where \[1hp = 746W\]
Complete answer:
Here we have asked to find a relation between distance \[\left( S \right)\] and time \[\left( t \right)\] in constant Power \[P\] .
The relation between Power, Force, and Velocity is,
\[P = F \times V\] …………..(1) Where \[P = \] Power; \[F = \] Force; \[V = \] Velocity
Let us divide the equation (1) into two-part as, one of the Force part and the other of the Velocity part. Coming to the Force part, let us write the equation of force as,
\[F = m \times a\]………(2), where \[m = \] mass and \[a = \] acceleration. And acceleration can be written as,
\[a = \dfrac{V}{t}\]…….(3) and \[V = \dfrac{S}{t}\]…….(4), where \[V = \]Velocity; \[S = \] distance, and \[t = \] time.
Comparing equations (3) and (4) it can be written as,
\[a = \dfrac{S}{{{t^2}}}\]………..(5). Substituting (5) in (2) we get,
\[F = m \times \dfrac{S}{{{t^2}}}\]………(6). Rewriting equation (1) in terms of (4) and (6),
\[P = m \times \dfrac{S}{{{t^2}}} \times \dfrac{S}{t}\]
\[ \Rightarrow Pm{t^3} = {S^2}\].
As Power and mass are constants,
\[ \Rightarrow {S^2} \propto {t^3}\]
That square of distance is proportional to the cube of the distance at constant Power.
Hence option (A) is correct.
Note: Power is a scalar quantity. That is, it has no direction but the only magnitude. Hence it is the scalar product of Force and Velocity.
At constant Power, \[V = \sqrt t \] where \[V = \] velocity and \[t = \]time. ( Can be derived from above equations)
Unit of Power is Watt(W). But horsepower unit is a standard measure for describing the power capabilities of daily equipment, where \[1hp = 746W\]
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