Question:easy

A particle moves with velocity $v = at - bt^2$ where a and b are constants. The acceleration becomes zero at:

Show Hint

Differentiate velocity once to get acceleration, then solve for t when it equals zero.
Updated On: Jun 10, 2026
  • $t = \frac{a}{b}$
  • $t = \frac{a}{2b}$
  • $t = \frac{2a}{b}$
  • $t = \frac{b}{a}$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Read the velocity rule.
The velocity of the particle is $v = a t - b t^{2}$, where $a$ and $b$ are constants. We must find the time when the acceleration becomes zero.

Step 2: Link acceleration to velocity.
Acceleration is how fast velocity changes with time. In calculus terms it is the derivative of velocity: $\text{acc} = \dfrac{dv}{dt}$.

Step 3: Differentiate each term.
The derivative of $a t$ is $a$. The derivative of $b t^{2}$ is $2 b t$. So the acceleration is $\text{acc} = a - 2 b t$.

Step 4: Apply the condition.
We want the moment when acceleration is zero, so set $a - 2 b t = 0$.

Step 5: Solve for time.
Rearranging gives $2 b t = a$, so $t = \dfrac{a}{2b}$.

Step 6: Check the meaning.
At this time the velocity reaches its peak, because acceleration switching from positive to negative marks the highest speed. \[ \boxed{t = \dfrac{a}{2b}} \]
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