
A constant power is delivered to a body moving along a straight line. The distance travelled by the body in time t is proportional to
A ${t^{1/2}}$
B ${t^{3/2}}$
C ${t^{5/2}}$
D ${t^{7/2}}$
Answer
220.5k+ views
Hint The above problem can be solved by using the concept of mechanical power. The power is the same as the work done by the object in unit time. It is also the same as the product of the force and velocity of the particle in the same direction.
Complete step by step answer
The power of the body in the above problem is constant.
The formula to calculate the power of the body is given as,
$P = F \cdot v......\left( 1 \right)$
Here, P is the power of the body
F is the force on the body
v is the speed of the body
The velocity of the particle is given as,
$v = \dfrac{s}{t}$
Here, s is the distance travelled by the body and t is the time in which the body covers the distance.
The force on the body is given as,
$F = m \cdot a......\left( 2 \right)$
Here, m is the mass of the body and a is the acceleration of the body.
The acceleration of the body is given as,
$a = \dfrac{s}{{{t^2}}}$
Substitute $\dfrac{s}{{{t^2}}}$for a in the equation (2) to find the force on the body.
$F = m\left( {\dfrac{s}{{{t^2}}}} \right)$
$F = \dfrac{{ms}}{{{t^2}}}$
Substitute $\dfrac{{ms}}{{{t^2}}}$for F and $\dfrac{s}{t}$for v in the equation (1) to find the power of the body.
$P = \left( {\dfrac{{ms}}{{{t^2}}}} \right)\left( {\dfrac{s}{t}} \right)$
$P = \dfrac{{m{s^2}}}{{{t^3}}}$
${s^2} = \left( {\dfrac{P}{m}} \right){t^3}$
$s = \sqrt {\dfrac{P}{m}} \left( {{t^{3/2}}} \right)......\left( 3 \right)$
In the above equation (3) the power and mass of the body is constant, so the distance covered by the body is proportional to the $3/2$th power of time.
Thus, the distance travelled by the body is proportional to ${t^{3/2}}$and the option (B) is the correct answer.
Note The power of the body can also be given by using the formula $P = \dfrac{W}{t}$ and work done by the body as $W = F \cdot s$. The above relation between the distance travelled by the body and time by $P = \dfrac{W}{t}$ and $W = F \cdot s$.
Complete step by step answer
The power of the body in the above problem is constant.
The formula to calculate the power of the body is given as,
$P = F \cdot v......\left( 1 \right)$
Here, P is the power of the body
F is the force on the body
v is the speed of the body
The velocity of the particle is given as,
$v = \dfrac{s}{t}$
Here, s is the distance travelled by the body and t is the time in which the body covers the distance.
The force on the body is given as,
$F = m \cdot a......\left( 2 \right)$
Here, m is the mass of the body and a is the acceleration of the body.
The acceleration of the body is given as,
$a = \dfrac{s}{{{t^2}}}$
Substitute $\dfrac{s}{{{t^2}}}$for a in the equation (2) to find the force on the body.
$F = m\left( {\dfrac{s}{{{t^2}}}} \right)$
$F = \dfrac{{ms}}{{{t^2}}}$
Substitute $\dfrac{{ms}}{{{t^2}}}$for F and $\dfrac{s}{t}$for v in the equation (1) to find the power of the body.
$P = \left( {\dfrac{{ms}}{{{t^2}}}} \right)\left( {\dfrac{s}{t}} \right)$
$P = \dfrac{{m{s^2}}}{{{t^3}}}$
${s^2} = \left( {\dfrac{P}{m}} \right){t^3}$
$s = \sqrt {\dfrac{P}{m}} \left( {{t^{3/2}}} \right)......\left( 3 \right)$
In the above equation (3) the power and mass of the body is constant, so the distance covered by the body is proportional to the $3/2$th power of time.
Thus, the distance travelled by the body is proportional to ${t^{3/2}}$and the option (B) is the correct answer.
Note The power of the body can also be given by using the formula $P = \dfrac{W}{t}$ and work done by the body as $W = F \cdot s$. The above relation between the distance travelled by the body and time by $P = \dfrac{W}{t}$ and $W = F \cdot s$.
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