A particle moving along x -axis has acceleration f, at time t, given by f = f 0(1-( t / T )), 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 ( v x ) is:

Question

A particle moving along x -axis has acceleration f, at time t, given by f = f 0(1-( t / T )), 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 ( v x ) is: (A) (1/2)f0 T2 (B) f0 T2 (C) (1/2)f0 T (D) f0 T

Tardigrade Admin · Accepted Answer

Given: At time t=0, velocity, v=0. Acceleration f=f0(1-(t/T)) At f=0, 0=f0(1-(t/T)) Since f0 is a constant, ∴ 1-(t/T)=0 or t=T text . Also, acceleration f=(d v/d t) ∴ ∫ limits0vx d v=∫ limitst=0t=T f d t=∫ limits0T f0(1-(t/T)) d t ∴ vx=[f0 t-(f0 t2/2 T)]0T =f0 T-(f0 T2/2 T)=(1/2) f0 T

Q. A particle moving along $x$ -axis has acceleration $f$ , at time $t$ , given by $f = f_{0} (1 - \frac{t}{T})$ , 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 $(v_{x})$ is:

$\frac{1}{2} f_{0} T^{2}$

$f_{0} T^{2}$

$\frac{1}{2} f_{0} T$

$f_{0} T$

Q. A particle moving along x -axis has acceleration f, at time t, given by f=f0​(1−Tt​), 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:

21​f0​T2

f0​T2

21​f0​T

f0​T

$\frac{1}{2} f_{0} T^{2}$

$f_{0} T^{2}$

$\frac{1}{2} f_{0} T$

$f_{0} T$