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Step-by-Step Solution
Step 1: Write down the given acceleration
The acceleration of the particle at time $t$ along the $x$-axis is given by
$$f(t) = f_0 \left(1 - \frac{t}{T}\right),$$
where $f_0$ and $T$ are constants.
Step 2: Determine the time when the acceleration becomes zero
Set $f(t) = 0$ to find the instant $t$ at which acceleration vanishes:
$$f_0 \left(1 - \frac{t}{T}\right) = 0 \quad \Longrightarrow \quad 1 - \frac{t}{T} = 0 \quad \Longrightarrow \quad t = T.$$
Step 3: Relate acceleration to velocity
Recall that acceleration is the time-derivative of velocity,
$$f(t) = \frac{dv_x}{dt}.$$
Thus, we can write
$$dv_x = f(t)\,dt.$$
Step 4: Integrate to find velocity
Since the velocity at $t=0$ is zero, integrate from $t=0$ to $t=T$ to find $v_x$:
$$ v_x = \int_{0}^{v_x} dv_x = \int_{0}^{T} f_0 \left(1 - \frac{t}{T}\right)\,dt.$$
Step 5: Evaluate the integral
1. First, distribute $f_0$ inside the parentheses:
$$ v_x = f_0 \int_{0}^{T} \left(1 - \frac{t}{T}\right)\,dt. $$
2. Split the integral:
$$ v_x = f_0 \left[\int_{0}^{T} 1\,dt - \int_{0}^{T} \frac{t}{T}\,dt\right].$$
3. Evaluate each integral:
- $\displaystyle \int_{0}^{T} 1\,dt = T.$
- $\displaystyle \int_{0}^{T} \frac{t}{T}\,dt = \frac{1}{T} \int_{0}^{T} t\,dt = \frac{1}{T} \cdot \frac{T^2}{2} = \frac{T}{2}.$
4. Combine the results:
$$ v_x = f_0 \left[T - \frac{T}{2}\right] = f_0 \cdot \frac{T}{2} = \frac{1}{2}\,f_0\,T.$$
Step 6: State the final velocity
Therefore, the velocity of the particle at the instant when the acceleration becomes zero is
$$ v_x = \frac{1}{2} f_0 T. $$
