Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A particle moves along a straight line OX. At a time t (in seconds), the displacement 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?

24 m

40 m

56 m

16 m

Step-by-Step Solution

Find Velocity ( $v$ ): Velocity is the rate of change of position with respect to time ( $v = dx/dt$ ) . Given $x = 40 + 12t - t^3$ . $v = \frac{d}{dt}(40 + 12t - t^3) = 12 - 3t^2$
Determine when the particle rests: The particle comes to rest when its velocity is zero. $12 - 3t^2 = 0$ $3t^2 = 12 \implies t^2 = 4 \implies t = 2 \text{ s}$
Calculate Positions: Initial position at $t = 0$ : $x(0) = 40 + 12(0) - 0^3 = 40$ m. Position at rest $t = 2$ : $x(2) = 40 + 12(2) - 2^3 = 40 + 24 - 8 = 56$ m.
Calculate Distance: Since the particle moves in a straight line without reversing direction until $t=2$ (velocity is positive for $t < 2$ ), the distance travelled is the magnitude of the displacement. $\text{Distance} = |x(2) - x(0)| = |56 - 40| = 16 \text{ m}$

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