Question

The position '$x$' of a particle varies with a time as $x = at^2 - bt^3$ where '$a$' and '$b$' are constants. The acceleration of the particle will be zero at}

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

Hint:

Acceleration is the second derivative of position ($d^2x/dt^2$). Always differentiate carefully with power rule.

Answer 1

The Correct Option is A

Concept:
Acceleration is the rate of change of velocity with time. If position of particle is given as: \[ x=f(t) \] then: \[ v=\frac{dx}{dt} \] \[ a=\frac{dv}{dt}=\frac{d^2x}{dt^2} \] We need the time when acceleration becomes zero.
Step 1: Given position equation
\[ x=at^2-bt^3 \] where $a$ and $b$ are constants.
Step 2: Find velocity
Differentiate position with respect to time: \[ v=\frac{dx}{dt} \] \[ v=\frac{d}{dt}(at^2-bt^3) \] \[ v=2at-3bt^2 \]
Step 3: Find acceleration
Differentiate velocity with respect to time: \[ a_{\text{cc=\frac{dv}{dt} \] \[ a_{\text{cc=\frac{d}{dt}(2at-3bt^2) \] \[ a_{\text{cc=2a-6bt \]
Step 4: Set acceleration equal to zero
For zero acceleration: \[ 2a-6bt=0 \] \[ 6bt=2a \] \[ t=\frac{2a}{6b} \] \[ t=\frac{a}{3b} \]
Step 5: Final Answer
The acceleration becomes zero at: \[ \boxed{\frac{a}{3b \] Quick Tip:
Whenever position is given, differentiate once for velocity and twice for acceleration.