Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

f₀T

1/2 f₀T²

f₀T²

1/2 f₀T

Step-by-Step Solution

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