A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as x = –3t3 + 18t2 + 16t, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m/s.
Correct Answer: 52
Approach Solution - 1
We are given the position equation:
\[ x = -3t^3 + 18t^2 + 16t \]
To find the velocity, we differentiate \(x(t)\) with respect to \(t\):
\[ v = \frac{dx}{dt} = -9t^2 + 36t + 16 \]
Next, we differentiate again to find the acceleration:
\[ a = \frac{dv}{dt} = -18t + 36 \]
Now, set the acceleration equal to zero to find the time when acceleration becomes zero:
\[ -18t + 36 = 0 \]
Solving for \(t\):
\[ t = 2 \, \text{s} \]
Substitute \(t = 2\) into the velocity equation:
\[ v = -9(2)^2 + 36(2) + 16 = -36 + 72 + 16 = 52 \, \text{m/s} \]
Thus, the velocity of the particle when its acceleration becomes zero is 52 m/s.
Approach Solution -2
The problem provides the position of a particle as a function of time, \( x(t) \), and asks for its velocity at the specific instant when its acceleration is zero.
Concept Used:
To solve this problem, we use the fundamental definitions of velocity and acceleration in kinematics, which are derived from the position function through differentiation.
Velocity (\(v\)): The instantaneous velocity is the first derivative of the position (\(x\)) with respect to time (\(t\)).
\[ v(t) = \frac{dx}{dt} \]Acceleration (\(a\)): The instantaneous acceleration is the first derivative of the velocity (\(v\)) with respect to time (\(t\)), which is also the second derivative of the position (\(x\)) with respect to time.
\[ a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} \]Step-by-Step Solution:
Step 1: Write down the given position function.
The position of the particle is given by the equation:
\[ x(t) = -3t^3 + 18t^2 + 16t \]Step 2: Differentiate the position function to find the velocity function, \(v(t)\).
Using the power rule for differentiation, \( \frac{d}{dt}(t^n) = nt^{n-1} \):
\[ v(t) = \frac{d}{dt}(-3t^3 + 18t^2 + 16t) \] \[ v(t) = -3(3t^2) + 18(2t) + 16(1) \] \[ v(t) = -9t^2 + 36t + 16 \]Step 3: Differentiate the velocity function to find the acceleration function, \(a(t)\).
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-9t^2 + 36t + 16) \] \[ a(t) = -9(2t) + 36(1) + 0 \] \[ a(t) = -18t + 36 \]Step 4: Find the time \(t\) when the acceleration becomes zero.
Set the acceleration function \(a(t)\) equal to zero and solve for \(t\):
\[ -18t + 36 = 0 \] \[ 18t = 36 \] \[ t = \frac{36}{18} = 2 \text{ s} \]So, the acceleration of the particle is zero at \(t = 2\) seconds.
Final Computation & Result:
Step 5: Substitute \(t = 2\) s into the velocity function \(v(t)\) to find the velocity at that instant.
\[ v(t) = -9t^2 + 36t + 16 \] \[ v(2) = -9(2)^2 + 36(2) + 16 \] \[ v(2) = -9(4) + 72 + 16 \] \[ v(2) = -36 + 72 + 16 \] \[ v(2) = 36 + 16 = 52 \text{ m/s} \]The velocity of the particle when its acceleration becomes zero is 52 m/s.
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