Motion in a Straight Line - Result Question 21
21. The position $x$ of a particle with respect to time $t$ along $x$-axis is given by $x=9 t^{2}-t^{3}$ where $x$ is in metres and $t$ in second. What will be the position of this particle when it achieves maximum speed along the +ve $x$ direction?
[2007]
(a) $54 m$
(b) $81 m$
(c) $24 m$
(d) $32 m$.
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Answer:
Correct Answer: 21. (a)
Solution:
- (a) Speed $v=\frac{d x}{d t}=\frac{d}{d t}(9 t^{2}-t^{3})=18 t-3 t^{2}$
For maximum speed its acceleration should be zero,
Acceleration $\frac{d v}{d t}=0 \Rightarrow 18-6 t=0 \Rightarrow t=3$
At $t=3$ its speed is max
$\Rightarrow x _{\text{max }}=81-27=54 m$
Position of any point is completely expressed by two factors its distance from the observer and its direction with respect to observer.
That is why position is characterised by a vector known as position vector.
22 .
(c) Here, $f=f_0(1-\frac{t}{T})$
or, $\frac{d v}{d t}=f_0(1-\frac{t}{T})$
If $f=0$, then
$0=f_0(1-\frac{t}{T}) \Rightarrow t=T$
Hence, particle’s velocity in the time interval $t=0$ and $t=T$ is given by
$v_x=\int _{v=0}^{v=V_z} d v=\int _{t=0}^{T}[f_0(1-\frac{t}{T})] d t$
$ \begin{aligned} & =f_0[(t-\frac{t^{2}}{2 T})]_0^{T} \\ & =f_0(T-\frac{T^{2}}{2 T})=f_0(T-\frac{T}{2})=\frac{1}{2} f_0 T . \end{aligned} $
