Question

The position x of a particle varies with time, (t) as $x = at^2 - bt^3.$ The acceleration will be zero at time t is equal to

• A. $\frac{a}{3b}$
• B. zero
• C. $\frac{2a}{3b}$
• D. $\frac{a}{b}$

Answer 1

The Correct Option is A

The problem asks us to determine the time \( t \) when the acceleration of a particle becomes zero, given its position as a function of time: \( x = at^2 - bt^3 \).

1. First, we find the velocity of the particle. The velocity \( v \) is the first derivative of position \( x \) with respect to time \( t \).

\[ v = \frac{dx}{dt} = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2 \]

2. Next, we find the acceleration \( a \) by differentiating the velocity \( v \) with respect to time \( t \).

\[ a = \frac{dv}{dt} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt \]

3. To find when the acceleration is zero, we set \( a = 0 \) and solve for \( t \).

\[ 2a - 6bt = 0 \quad \Rightarrow \quad 2a = 6bt \quad \Rightarrow \quad t = \frac{2a}{6b} = \frac{a}{3b} \]

Hence, the time at which the acceleration of the particle becomes zero is \( \frac{a}{3b} \).

• This matches the given correct option, which is \( \frac{a}{3b} \).