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A particle moves along a straight line O...

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

A

`24m`

B

`40m`

C

`56m`

D

`16m`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow the given information and derive the necessary equations. ### Step 1: Understand the given equation The position of the particle as a function of time is given by: \[ x(t) = 40 + 12t - t^3 \] ### Step 2: Find the velocity of the particle The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \): \[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(40 + 12t - t^3) \] Calculating the derivative: \[ v(t) = 0 + 12 - 3t^2 \] So, we have: \[ v(t) = 12 - 3t^2 \] ### Step 3: Determine when the particle comes to rest The particle comes to rest when its velocity is zero: \[ v(t) = 0 \] Setting the velocity equation to zero: \[ 12 - 3t^2 = 0 \] Solving for \( t \): \[ 3t^2 = 12 \] \[ t^2 = 4 \] Taking the square root: \[ t = \pm 2 \] Since time cannot be negative, we take: \[ t = 2 \, \text{seconds} \] ### Step 4: Calculate the position at \( t = 2 \) seconds Now, we need to find the position of the particle at \( t = 2 \): \[ x(2) = 40 + 12(2) - (2)^3 \] Calculating this: \[ x(2) = 40 + 24 - 8 \] \[ x(2) = 40 + 16 = 56 \, \text{meters} \] ### Step 5: Calculate the distance traveled before coming to rest The initial position at \( t = 0 \) is: \[ x(0) = 40 \, \text{meters} \] The distance traveled before coming to rest is: \[ \Delta x = x(2) - x(0) \] \[ \Delta x = 56 - 40 = 16 \, \text{meters} \] ### Final Answer The particle travels **16 meters** before coming to rest. ---
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