SATHEE: Kinematics Question 364

Kinematics Question 364

Question: A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from O it’s given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

Options:

A) 24 m

B) 40 m

C) 56 m

D) 16 m

Show Answer

Answer:

Correct Answer: C

Solution:

When particle comes to rest, $V=0= \frac{d x}{d t} = \frac{d}{d t} (40 + 12 t - t^{3})$
$\Rightarrow 12 - 3 t^{2} = 0$
$\Rightarrow t^{2} = \frac{12}{3} = 4 ∴ t=2 sec$

therefore distance travelled by particle before coming to rest, $x = 40 + 12 t - t^{3} = 40 + 12 \times 2 - {(2)}^{3} = 56 m$

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Kinematics Question 364

Question: A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from O it’s given by x=40+12t−t3 . How long would the particle travel before coming to rest?

Options:

Answer:

Solution:

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Question: A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from O it’s given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?