Question

The position of an object moving along a line is given by the function s(t)=−15t2+75t. Find the average velocity of the object over the following intervals.

​(d) ​[1, 1+​h] where h>0 is any real number.

Answer 1

$Solution: \\Method ~1: \\The ~ratio ~of ~the~ total~ displacement~ with~ respect ~to~ time ~is ~known ~as ~\\average ~velocity. If ~an ~object~ covers ~some~ displacement~ which ~is~ given~ by~ a~\\ function~ S(t) .The ~formula~ for~ average~ velocity~ over ~the~ interval~ [t_1,bt_2] ~is \\V_{average}=\frac{S(t_2)-S(t_1)}{t_2-t_1} \\ Given ~function~ S(t)=-15t^2+75t ~on ~the~ interval~[1,1+h], \\where~h>0~is~any ~real ~number. \\ The~ average~ velocity ~of~ the~ object~ over ~the~interval \\V_{average}=\frac{S(1+h)-S(1)}{(1+h)-1}=\frac{S(1+h)-S(1)}{h} \\S(1+h)=-15(1+h)^2 +75(1+h)=-15(1^2+2h+h^2) +(75+75h) \\~~~~~~~~~~~~~~~~~=-15(1+2h+h^2)+75+75h=-15-30h-15h^2+75+75h \\~~~~~~~~~~~~~~~~~=60+45h-15h^2=-15h^2+45h+60 \\\therefore S(1+h)=-15h^2+45h+60~~~and~~S(1)=-15(1)^2+75(1)=-15+75=60 \\ \therefore V_{average}=\frac{S(1+h)-S(1)}{h}=\frac{-15h^2+45h+60-60}{h}=\frac{-15h^2+45h}{h}=-15h+45 \\\therefore V_{average}=-15h+45$$Method ~~2: \\General ~equations~ for~ the~ position ~of ~the ~body ~and ~its ~velocity ~for ~uniformly~ \\ accelerated ~motion ~have ~the ~form: \\S(t)=v_0t+\frac{at^2}{2},~v(t)=v_0+at~,where~ v_0~is~intial ~velocity~ and ~a~is ~acceleration. \\from~the~condition~of~the~problem~it ~is~ seen~ that~ the ~initial ~velocity~ and~ \\acceleration ~are~ respectively~ equal ~to ~75 ~and ~-30. ~Then \\v(t)=75-30t \\The ~average ~velocity ~in ~the ~interval~ [t_1, t_2] ~for ~uniformly ~accelerated ~motion ~is\\ equal~ to ~the ~average ~value ~between ~the ~initial ~and ~final ~velocities ~for ~a ~given~ \\interval. \\v_{average}=\frac{v(t_1)+v(t_2)}{2}=\frac{(75-30t_1)+(75-30t_2)}{2}=\frac{150-30(t_1+t_2)}{2}=75-15(t_1+t_2) \\Here ~the ~interval ~[t_1,t_2]=[1,1+h] \\\therefore v_{average}=75-15(t_1+t_2)=75-15(1+(1+h))=75-15(2+h) \\~~~~~~~~~~~~~~~~~=75-30-15h=45-15h=-15h+45 \\\therefore v_{average}=-15h+45$