Question

If the distance travelled by a particle in \(t\) seconds is given by \(S = 72t + 3t^2 - t^3\), then the time taken by the particle to come to rest is:

• A. 4 seconds
• B. 6 seconds
• C. 3 seconds
• D. 0 seconds

Hint:

“Coming to rest” mathematically translates directly to setting the first derivative \(\frac{dS}{dt} = 0\). Always make sure to filter out negative time solutions at the end of quadratic calculations since time cannot run backwards.

Answer 1

The Correct Option is B

Concept: In kinematics, the position or displacement of an object is given as a function of time, \(S(t)\).
• The instantaneous velocity \(v(t)\) is defined as the first time-derivative of the displacement function: \[ v = \frac{dS}{dt} \]
• When an object is said to “come to rest”, its instantaneous linear velocity drops to exactly zero (\(v = 0\)).

Step 1: Compute the velocity function \(v\)
Given the displacement equation: \[ S = 72t + 3t^2 - t^3 \] Differentiating each term with respect to the independent variable \(t\) using the power rule (\(\frac{d}{dt}(t^n) = n t^{n-1}\)): \[ v = \frac{dS}{dt} = \frac{d}{dt}(72t) + \frac{d}{dt}(3t^2) - \frac{d}{dt}(t^3) \] \[ v = 72(1) + 3(2t) - 3t^2 \] \[ v = 72 + 6t - 3t^2 \]

Step 2: Set the velocity equation equal to zero
To find the exact timestamp when the particle is completely stationary (at rest), equate \(v\) to \(0\): \[ 72 + 6t - 3t^2 = 0 \] Let us rearrange this into a standard quadratic equation format (\(at^2 + bt + c = 0\)) by multiplying the entire expression by \(-1\): \[ 3t^2 - 6t - 72 = 0 \] Divide every term by the common coefficient factor of $3$ to simplify the quadratic factorization: \[ t^2 - 2t - 24 = 0 \]

Step 3: Factorize the quadratic equation
We need two numbers that multiply to give \(-24\) and add together to form \(-2\). These numbers are \(-6\) and \(+4\). Split the middle linear term: \[ t^2 - 6t + 4t - 24 = 0 \] Group terms by factoring out common values: \[ t(t - 6) + 4(t - 6) = 0 \] \[ (t - 6)(t + 4) = 0 \] This leaves us with two potential mathematical root solutions for the time variable: \[ t - 6 = 0 \quad \Rightarrow \quad t = 6 \text{ seconds} \] \[ t + 4 = 0 \quad \Rightarrow \quad t = -4 \text{ seconds} \]

Step 4: Filter out extraneous solutions
Since time elapsed cannot be a negative value in real physical situations (\(t \ge 0\)), we discard \(t = -4\). Therefore, the single valid time calculation is \(t = 6\) seconds.