Question

A particle moving along $x$ -axis has acceleration $f$, at time $t$, given by $f = f _{0}\left(1-\frac{ t }{ T }\right)$, where $f _{0}$ and $T$ are constants. The particle at $t =0$ has zero velocity. In the time interval between $t =0$ and the instant when $f =0$, the particle's velocity $\left( v _{ x }\right)$ is:

• A. $\frac{1}{2}f_0 T^2$
• B. $f_0 T^2$
• C. $\frac{1}{2}f_0 T$
• D. $f_0 T$

Answer 1

The Correct Option is C

 To solve this problem, we need to find the velocity of the particle when the acceleration becomes zero. Let's go through the problem step-by-step:

Given:

• Acceleration \( f = f_0 \left(1 - \frac{t}{T}\right) \)
• Initial velocity \( v_x(t=0) = 0 \)
• We need to find the velocity \( v_x \) at the instant \( t = T \) when \( f = 0 \).

 

Step 1: Find the time when acceleration is zero.

• Set \( f = 0 \) and solve for \( t \):
• \( f_0 \left(1 - \frac{t}{T}\right) = 0 \)
• This gives \( 1 - \frac{t}{T} = 0 \Rightarrow t = T \).

Step 2: Integrate acceleration to get velocity.

• The velocity is the integral of acceleration with respect to time:
• \( v_x = \int_0^T f \, dt = \int_0^T f_0 \left(1 - \frac{t}{T}\right) \, dt \)
• \( v_x = \int_0^T \left(f_0 - \frac{f_0}{T} t\right) \, dt \)
• This separates into: \( v_x = f_0 \int_0^T 1 \, dt - \frac{f_0}{T} \int_0^T t \, dt \)
• \( v_x = f_0 \left[t\right]_0^T - \frac{f_0}{T} \left[\frac{t^2}{2}\right]_0^T \)

Step 3: Evaluate the integrals.

• First term: \( f_0 [T - 0] = f_0 T \)
• Second term: \(- \frac{f_0}{T} \left[\frac{T^2}{2} - \frac{0^2}{2}\right] = - \frac{f_0}{T} \cdot \frac{T^2}{2} = - \frac{f_0 T}{2} \)

Step 4: Combine results to find velocity.

• Combining these results, we have:
• \( v_x = f_0 T - \frac{f_0 T}{2} = \frac{f_0 T}{2} \)

Conclusion:

• Thus, the velocity of the particle when the acceleration becomes zero is \( \frac{1}{2}f_0 T \).