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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Mass of the body, m = 2\,\text{kg}
• Power delivered by the engine, P = 1\,\text{J/s}
• Body starts from rest, so initial velocity u = 0\,\text{m/s}
• Time elapsed, t = 9\,\text{s}
Step 2: Relate Power and Change in Kinetic Energy
From the definition of power,
P = \frac{\text{Work done}}{\text{Time}} = \frac{\Delta K}{t},
where \Delta K is the change in kinetic energy. Hence,
P\,t = \Delta K = \frac{1}{2} m v^2.
Substituting the values P=1\,\text{J/s} , t=9\,\text{s} , and m=2\,\text{kg} ,
1 \times 9 = \frac{1}{2} \times 2 \times v^2.
Simplifying, we get
9 = v^2 \quad \Rightarrow \quad v = 3\,\text{m/s}.
This is the velocity of the body at the end of 9\,\text{s} .
Step 3: Express Power in Terms of Force and Velocity
We know P = F \, v. Since F = ma , we can write
P = ma \, v = m \left(\frac{dv}{dt}\right) v.
But v = \frac{ds}{dt} , so a = \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v \frac{dv}{ds}.
Substituting this back gives
P = m \left(v \frac{dv}{ds}\right) v = m \, v^2 \frac{dv}{ds}.
Rearranging to integrate with respect to v and s :
m \, v^2 \frac{dv}{ds} = P \quad \Rightarrow \quad v^2 dv = \frac{P}{m} \, ds.
Since P=1\,\text{J/s} and m=2\,\text{kg} , the above becomes
v^2 dv = \frac{1}{2} \, ds.
Step 4: Integrate to Find the Distance
We integrate both sides from the initial velocity 0\,\text{m/s} to the final velocity 3\,\text{m/s} , and from s=0 to s=S , the distance traveled in that period:
\int_{0}^{3} v^2 \, dv \;=\; \frac{1}{2} \int_{0}^{S} ds.
Carry out the left side integration:
\int_{0}^{3} v^2 \, dv = \left[ \frac{v^3}{3} \right]_{0}^{3} = \frac{3^3}{3} = \frac{27}{3} = 9.
So the equation becomes
9 = \frac{1}{2} \cdot S.
Solving for S , we get
S = 18\,\text{m}.
Step 5: Final Answer
The distance covered by the body in 9\,\text{s} is 18\,\text{m}.
Solution Reference Image
