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:

  1. (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} $