Question

* Answer the question with steps *

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Answer 1

We are given a relation between acceleration (f) and time (t):


$\boxed{f=f_{\circ}\left( 1- \frac{t}{T} \right)} \quad ---(1)$

$f_{\circ}$ and $T$ are given as constants. 



The question asks about the particle's velocity when f=0. So we must find some equation of velocity. Here it is the easiest to find an equation of velocity as a function of time.

The particle is moving along X-axis. So its velocity is being denoted as $v_x$


Now, we know that acceleration is the rate of change of velocity with respect to time. In Calculus terms, acceleration is the derivative of velocity with respect to time.



We write:

$\displaystyle f = \frac{dv_x}{dt} \\ \\ \\ \implies dv_x = f \, dt \\ \\ \\ \implies v_x = \int f \, dt \\ \\ \\ \implies v_x = \int f_{\circ} \left( 1-\frac{t}{T} \right) \, dt \\ \\ \\ \implies v_x = f_{\circ} \, t - \frac{f_{\circ}}{T} \times \frac{t^2}{2} + c \\ \\ \\ \implies v_x = f_{\circ}t - \frac{f_{\circ}t^2}{2T}+c \\ \\ \\ \implies v_x = f_{\circ}t \left( 1 - \frac{t}{2T} \right) +c$

$c$ is the constant of integration. We need to find its value. 


In the question, we are given that The particle has zero velocity at t = 0


So we can put $v_x=0$ and $t=0$ to find c.


$\displaystyle v_x = f_{\circ}t \left( 1 - \frac{t}{2T} \right) +c \\ \\ \\ \implies 0 = f_{\circ} \times 0 \times \left( 1-\frac{t}{2T} \right) +c \\ \\ \\ \implies 0 = 0 + c \\ \\ \\ \implies \bold{c = 0}$


So, our equation of velocity becomes:


$\displaystyle \boxed{\bold{v_x = f_{\circ}t \left( 1 - \frac{t}{2T} \right)}}$



Now, we have to find the velocity when f = 0.


But we have velocity as a function of t, and not f. So, we use equation (1).


$f = f_{\circ} \left(1-\frac{t}{T} \right) \\ \\ \\ \text{Put }f=0 \\ \\ \\ \implies 0 = f_{\circ} \left( 1-\frac{t}{T} \right) \\ \\ \\ \implies 1-\frac{t}{T}=0 \\ \\ \\ \implies \frac{t}{T}=1 \\ \\ \\ \implies \bold{t = T}$


Thus, we know that acceleration f becomes zero at time t = T. We can put this value of t in our equation to find velocity.


$\displaystyle v_x = f_{\circ}t \left( 1 - \frac{t}{2T} \right) \\ \\ \\ \text{Put }t=T \\ \\ \\ \implies v_x = f_{\circ}T \left( 1-\frac{T}{2T} \right) \\ \\ \\ \implies v_x = f_{\circ}T \left( 1-\frac{1}{2} \right) \\ \\ \\ \implies v_x = f_{\circ}T \times \frac{1}{2} \\ \\ \\ \implies \boxed{\bold{v_x=\frac{1}{2}f_{\circ}T}}$


Thus, when f = 0, the velocity of the particle is $\frac{1}{2}f_{\circ}T$



So, The Answer is Option (1)