Motion In A Straight Line Ques 1
1. A particle moving along $x$-axis has acceleration $f$, at time $t$, given by $f=f_0\left(1-\frac{t}{T}\right)$, where $\mathrm{f}_0$ and $\mathrm{T}$ 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 $\left(v_x\right)$ is
[2007]
(a) $\frac{1}{2} f_0 \mathrm{T}^2$
(b) $f_0 \mathrm{T}^2$
(c) $\frac{1}{2} f_0 \mathrm{T}$
(d) $f_0 \mathrm{T}$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Here, $f=f_0\left(1-\frac{t}{T}\right)$
or, $\frac{d v}{d t}=f_0\left(1-\frac{t}{T}\right)$
If $f=0$, then
$ 0=f_0\left(1-\frac{t}{T}\right) \Rightarrow t=T $
Hence, particle’s velocity in the time interval $t=0$ and $t=T$ is given by
$ v_x=\int_{v=0}^{v=V_2} d v=\int_{t=0}^T\left[f_0\left(1-\frac{t}{T}\right)\right] d t $
$ \begin{aligned} & =f_0\left[\left(t-\frac{t^2}{2 T}\right)\right]_0^T \ & =f_0\left(T-\frac{T^2}{2 T}\right)=f_0\left(T-\frac{T}{2}\right)=\frac{1}{2} f_0 T . \end{aligned} $
BETA
