Question

A body is projected from the origin 0 and moves in a straight line such that its displacement from 9 , t seconds after projection, is s metres where s=t^3/3 - 6t^2 + 50t. The velocity of the body , u m/sec., t seconds after projection is given by ds/dt. ? For what value of t is the body moving with minimum velocity and how far from 0 is the body at this time?

Answer 1

WELCOME TO OPEN STUDY

Answer 2

minimum and maximum application

Answer 3

just differentiate it

Answer 4

s=t^3/3 - 6t^2 + 50t

Answer 5

ds/dt = v

Answer 6

do a 2nd derivative to check the minimum

Answer 7

if u take the second derivatite it will be acceleration

Answer 8

if f'' x>0 then its the minimum

Answer 9

Thank you all! I ve tried to dif many times, just cannot get the correct answer sorry (such that its displacement from 0)
t shoulb be 6!

Answer 10

Once again, thanks

Answer 11

I ve solved T, its really 6!