Question

A car of mass $m$ moving with velocity $u$ on a straight road in a straight line, doubles its velocity in time $t$. The power delivered by the engine of a car for doubling the velocity is

• A. $\frac{3mu^2}{2t}$
• B. $\frac{mu^2}{2t}$
• C. $\frac{2mu^2}{t}$
• D. $\frac{3mu^2}{t}$

Hint:

You can use a quick algebraic shortcut: $\Delta K.E. = \frac{1}{2}m(v_f^2 - v_i^2)$.
Substituting $v_f = 2u$ and $v_i = u$ directly gives: $v_f^2 - v_i^2 = 4u^2 - u^2 = 3u^2$.
Plugging this into the power formula yields $P = \frac{3mu^2}{2t}$ in a single step!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Calculate the power delivered during acceleration using a direct algebraic shortcut on the kinetic energy change.

Step 2: Key Formula or Approach:
Change in kinetic energy ΔK.E. = ½m(v_f² - v_i²). Average power P = ΔK.E./t. Substitute final and initial velocities directly without intermediate steps.

Step 3: Detailed Explanation:
Given v_f = 2u and v_i = u, compute v_f² - v_i² = (2u)² - u² = 4u² - u² = 3u². Plugging into the kinetic energy expression: ΔK.E. = ½m(3u²) = (3mu²)/2. Dividing by time t gives the average power: P = (3mu²)/(2t). This streamlined substitution bypasses separate calculations of initial and final kinetic energies.

Step 4: Final Answer:
The average power equals 3mu²/(2t).