Question

A particle moving from a fixed point on a straight line travels a distance \(S\) meters in \(t\) sec. If \(S = t^3 - t^2 + t + 3\), then the distance (in mts) travelled by the particle when it comes to rest is:

• A. \(5\)
• B. \(4\)
• C. \(2\)
• D. \(3\)

Hint:

When calculating real-world values, keep track of units and carefully perform substitution and simplification.

Answer 1

The Correct Option is C

Given the distance function: \[ S(t) = t^3 - t^2 - t + 3 \] The velocity function is the derivative of the distance function: \[ v(t) = \frac{dS}{dt} = 3t^2 - 2t - 1 \] The particle comes to rest when the velocity is zero, so we set \(v(t) = 0\): \[ 3t^2 - 2t - 1 = 0 \] We factor the quadratic equation: \[ (3t + 1)(t - 1) = 0 \] This gives us two possible times: \[ t = -\frac{1}{3} \quad \text{or} \quad t = 1 \] Since time cannot be negative, we have \(t = 1\) second. Now, we find the distance traveled at \(t = 1\) second by substituting \(t = 1\) into the distance function: \[ S(1) = (1)^3 - (1)^2 - (1) + 3 \] \[ S(1) = 1 - 1 - 1 + 3 \] \[ S(1) = 2 \] Therefore, the distance traveled by the particle when it comes to rest is 2 meters. Answer: 2