A particle moving along the x-axis has acceleration f, at time t, given by f = f₀(1 - t/T), where f₀ and T are

Question

A particle moving along the x-axis has acceleration f, at time t, given by f = f₀(1 - t/T), where f₀ 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 (vₓ) is:

Accepted Answer

1.  **Identify the condition for the time interval:** The interval ends when the acceleration $f$ becomes zero.
    $$ f = f_0 \left( 1 - \frac{t}{T} \right) = 0 $$
    $$ 1 - \frac{t}{T} = 0 \implies t = T $$
2.  **Relate acceleration and velocity:** Acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$) .
    $$ dv = f \, dt = f_0 \left( 1 - \frac{t}{T} \right) dt $$
3.  **Integrate to find velocity:** Integrate from $t=0$ (where $v=0$) to $t=T$.
    $$ v_x = \int_{0}^{T} f_0 \left( 1 - \frac{t}{T} \right) dt $$
    $$ v_x = f_0 \left[ t - \frac{t^2}{2T} \right]_{0}^{T} $$
4.  **Evaluate the limits:**
    $$ v_x = f_0 \left( T - \frac{T^2}{2T} \right) - 0 $$
    $$ v_x = f_0 \left( T - \frac{T}{2} \right) = \frac{1}{2} f_0 T $$

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