Question

A particle moving in a straight line starts from rest, and the acceleration at any time $t$ is $a - kt^2$, where $a$ and $k$ are positive constants. The maximum velocity attained by the particle is

• A. 2/3√a3/k
• B. 1/3√a3/k
• C. √a3/k
• D. 2√a3/k

Answer 1

The Correct Option is A

The velocity of the particle increases as long as the acceleration is positive, and starts decreasing when acceleration becomes negative.

Step 1: Find the time \( t \) when velocity is maximum
The velocity is maximum when acceleration becomes zero: \[ a(t) = a - kt^2 = 0 \Rightarrow t^2 = \frac{a}{k} \Rightarrow t = \sqrt{\frac{a}{k}} \]

Step 2: Integrate acceleration to get velocity
We know that: \[ v(t) = \int a(t) \, dt = \int (a - kt^2) \, dt = at - \frac{k}{3} t^3 + C \] Since the particle starts from rest: \[ v(0) = 0 \Rightarrow C = 0 \Rightarrow v(t) = at - \frac{k}{3} t^3 \]

Step 3: Find maximum velocity
Substitute \( t = \sqrt{\frac{a}{k}} \) into the expression for \( v(t) \):

\[ v_{\text{max}} = a \cdot \sqrt{\frac{a}{k}} - \frac{k}{3} \left( \sqrt{\frac{a}{k}} \right)^3 = a \cdot \sqrt{\frac{a}{k}} - \frac{k}{3} \cdot \frac{a^{3/2}}{k^{3/2}} \]

Simplify: \[ v_{\text{max}} = \frac{a^{3/2}}{k^{1/2}} - \frac{1}{3} \cdot \frac{a^{3/2}}{k^{1/2}} = \left(1 - \frac{1}{3}\right) \cdot \frac{a^{3/2}}{k^{1/2}} = \frac{2}{3} \cdot \frac{a^{3/2}}{k^{1/2}} \]

Final Answer: \[ \boxed{\frac{2}{3} \cdot \frac{a^{3/2}}{\sqrt{k}}} \]