SATHEE: Motion in a Straight Line

Motion in a Straight Line - Result Question 22

23. A particle moves along a straight line OX. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from O is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

[2006]

(a) $40 m$

(b) $56 m$

(d) $24 m$

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Answer:

Correct Answer: 23. (b)

Solution:

$\begin{matrix} (d) & (b) x = 40 + 12 t - t^{3} \end{matrix}$

$v = \frac{d x}{d t} = 12 - 3 t^{2}$

For $v = 0; t = \sqrt{\frac{12}{3}} = 2 s e c$

So, after 2 seconds velocity becomes zero.

Value of $x$ in 2 secs $= 40 + 12 \times 2 - 2^{3}$

$= 40 + 24 - 8 = 56 m$

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Motion in a Straight Line - Result Question 22

23. A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x=40+12t−t3. How long would the particle travel before coming to rest?

Answer:

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23. A particle moves along a straight line OX. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from O is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?