Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Laws of Motion

Question:

A particle moving along x-axis has acceleration f, at time t, given by \(f = f_o[1-\frac{t}{T}]\) , 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 : 

Options:

\(\frac{1}{2}f_oT^2\)

\(f_oT^2\)

\(\frac{1}{2}f_oT\)

\(f_oT\)

Correct Answer:

\(\frac{1}{2}f_oT\)

Explanation:

Acceleration : \(f = f_o[1-\frac{t}{T}] = \frac{dv}{dt}\)

integrating : \(\int dv = \int  f_o[1-\frac{t}{T}] dt\) 

\( v = f_o t - \frac{t^2}{2T} + C \)

Initial condition : v(t = o) = 0

\( 0 = 0 - 0 + C\) ⇒ C = 0

Thus, velocity at any time t : \( v(t) = f_o t - \frac{t^2}{2T} \)

when f = 0 ⇒ \(0 = f_o[1-\frac{t}{T}] \)

Thus, t = T

\( v(t) = f_o T - \frac{T^2}{2T} = \frac{f_0}{2T}\)