The space s described in time t by a particle moving in a straight line is given by s=t^5-40t^3+30t^2+80t-250. Find the minimum value of its acceleration.

Asked by shaktidevgun | 28th May, 2016, 07:10: PM

Expert Answer:

s equals t to the power of 5 minus 40 t cubed plus 30 t squared plus 80 t minus 250
v equals fraction numerator d s over denominator d t end fraction equals 5 t to the power of 4 minus 120 t squared plus 60 t plus 80
a equals fraction numerator d v over denominator d t end fraction equals 20 t cubed minus 240 t plus 60
F o r space e x t r e m a comma space fraction numerator d a over denominator d t end fraction equals 0
rightwards double arrow 60 t squared minus 240 equals 0
rightwards double arrow t squared equals 4 rightwards double arrow t equals plus-or-minus 2
fraction numerator d a over denominator d t end fraction equals 60 open parentheses t plus 2 close parentheses open parentheses t minus 2 close parentheses
fraction numerator d a over denominator d t end fraction space c h a n g e s space s i g n space f r o m space n e g a t i v e space t o space p o s i t i v e space a t space t equals 2.
H e n c e comma space b y space f i r s t space d e r i v a t i v e space t e s t space a c c e l e r a t i o n space i s space m i n i m u m space a t space t equals 2
M i n i m u m space a c c e l e r a t i o n equals 20 cross times 2 cubed minus 240 cross times 2 plus 60
equals 160 minus 480 plus 60
equals negative 260

Answered by satyajit samal | 29th May, 2016, 05:39: PM