Question 39

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Jambmaths question: 

The distance travelled by a particle from a fixed point is given as $s={{t}^{3}}-{{t}^{2}}-t+5$find the minimum distance that the particle can cover from the fixed point.

Jamb Maths Solution: 

$\begin{align}  & s={{t}^{3}}-{{t}^{2}}-t+5 \\ & \frac{ds}{dt}=3{{t}^{2}}-2t-1 \\ & \text{At stationary point}\frac{dy}{dx}=0 \\ & 3{{t}^{2}}-2t-1=0 \\ & (3t+1)(t-1)=0 \\ & t=-\tfrac{1}{3}\text{ or }t=1 \\ & \frac{{{d}^{2}}s}{d{{t}^{2}}}=6t-2 \\ & \text{ignore }t=-\tfrac{1}{3}\text{ because }t\text{ cannot be negative } \\ & \text{when }t=1 \\ & \frac{{{d}^{2}}s}{d{{t}^{2}}}=6(1)-2=4>0\text{  (min)} \\ & \text{when }t=1\text{ will yield minimum distance} \\ & {{S}_{\min }}={{1}^{3}}-{{1}^{2}}-1+5=4cm \\\end{align}$

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