Question

A body undergoing a straight line motion is under the influence of source of constant power. How does its displacement very with time?

• A. 20 m
• B. 40 m
• C. 10\(\sqrt{2}\) m
• D. \( 20\sqrt{2}\)
• E. 5\(\sqrt{10}\) m

Answer 1

The Correct Option is D

Given:

• Body moving under constant power (P = constant)
• Displacement at t = 0 s: 0 m
• Displacement at t = 10 s: 10 m

Step 1: Relate Power to Motion

For constant power, the work done (W) is proportional to time (t):

\[ W = Pt \]

Work translates to kinetic energy: \( W = \frac{1}{2}mv^2 \). Thus, velocity (v) depends on time as:

\[ v = \sqrt{\frac{2Pt}{m}} \propto \sqrt{t} \]

Step 2: Integrate Velocity to Find Displacement

Displacement (s) is the integral of velocity:

\[ s(t) = \int_0^t v \, dt \propto \int_0^t \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \]

Thus, displacement follows:

\[ s(t) = k t^{3/2} \]

Step 3: Determine Proportionality Constant (k)

Given \( s(10) = 10 \, \text{m} \):

\[ 10 = k \cdot 10^{3/2} \implies k = \frac{10}{10^{3/2}} = 10^{-1/2} \]

Step 4: Calculate Displacement at t = 20 s

\[ s(20) = 10^{-1/2} \cdot 20^{3/2} = \frac{20^{3/2}}{10^{1/2}} = \left(\frac{20^3}{10}\right)^{1/2} = (800)^{1/2} = 20\sqrt{2} \, \text{m} \]

Conclusion:

The displacement at \( t = 20 \, \text{s} \) is \( 20\sqrt{2} \, \text{m} \).

Answer: \(\boxed{D}\)

Answer 2

1. Understand the relationship between power, work, and displacement:

Power (P) is the rate at which work (W) is done:

\[P = \frac{W}{t}\]

Work done is equal to the change in kinetic energy (KE):

\[W = \Delta KE\]

Kinetic energy is related to mass (m) and velocity (v):

\[KE = \frac{1}{2}mv^2\]

For constant power, we also have:

\[Pt = \frac{1}{2}mv^2\]

And displacement (s) is related to velocity (v) and time(t) under constant acceleration (which would arise from constant power on the same body):

\[ s = s_0 + v_0t + \frac{1}{2}at^2 \]

2. Define variables and given information:

• P = constant (power)
• At t = 0, s = 0
• At t = 10 s, s = 10 m
• Find displacement at t = 20 s

3. Derive the displacement equation for constant power:

Since power is constant, \(Pt = \frac{1}{2}mv^2\). We can express v in terms of P and t: \[v = \sqrt{\frac{2Pt}{m}} \]

For motion under constant acceleration and starting from rest at s=0, integrating velocity gives us the displacement: \[s = s_0 + v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2\] For constant power we can write: \[P = Fv = mav\] \[P = ma \sqrt{\frac{2Pt}{m}}\] \[a = \frac{\sqrt{2P}}{2m^{3/2} \times \sqrt{t}} = \frac{k}{\sqrt{t}}\] where k is constant since P and m are constant.

\[s = s_0 + \frac{1}{2}at^2 = \frac{1}{2}kt^{3/2}\] where k is constant since P and m are constant.

 

4. Use the given information to solve for the constant k:

At t = 10 s, s = 10 m:

\[10 = \frac{1}{2} k (10)^{3/2}\]

\[k = \frac{20}{10\sqrt{10}} = \frac{2}{\sqrt{10}}\]

5. Calculate the displacement at t = 20 s:

\[s = \frac{1}{2} (\frac{2}{\sqrt{10}}) (20)^{3/2} = \frac{1}{\sqrt{10}} \times 20\sqrt{20} = 20 \times \sqrt{2} = 20\sqrt{2} \, m\]

Final Answer: The final answer is \(\boxed{D}\)