The displacement, \( s \) metres, of an object is described by the equation \( s=2 t^{3}-18 t^{2}+96 t \), where \( t \) is the time in minutes and \( t \geq 0 \). Find the time at which the object has the smallest velocity. If your answer is a decimal, give it to 1 d.p.
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution The Deep Dive
To find the time at which the object has the smallest velocity, we first need to find the velocity function by differentiating the displacement function with respect to time \( t \). 1. **Differentiate the displacement function**: \[ s = 2t^3 - 18t^2 + 96t \] The velocity \( v \) is the first derivative of \( s \): \[ v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 18t^2 + 96t) = 6t^2 - 36t + 96 \] 2. **Set the velocity function to zero to find critical points**: To find the minimum velocity, we set the derivative equal to zero: \[ 6t^2 - 36t + 96 = 0 \] Dividing the entire equation by 6 simplifies it: \[ t^2 - 6t + 16 = 0 \] 3. **Apply the quadratic formula**: Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -6 \), and \( c = 16 \): \[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \] \[ t = \frac{6 \pm \sqrt{36 - 64}}{2} \] \[ t = \frac{6 \pm \sqrt{-28}}{2} \] Since the discriminant is negative, this means there are no real roots. 4. **Analyze the behavior of the velocity function**: Since the coefficient of \( t^2 \) in the velocity function \( 6t^2 - 36t + 96 \) is positive, the function opens upwards, meaning it has a minimum point. 5. **Find the vertex**: The time at which the minimum occurs can be found using the vertex formula \( t = -\frac{b}{2a} \): \[ t = -\frac{-36}{2 \cdot 6} = \frac{36}{12} = 3 \] Thus, the time at which the object has the smallest velocity is \( \boxed{3.0} \) minutes.
