4 votes 4 votes Velocity of an object fired directly in upward direction is given by ܸ$V\mathit{}=80-32 t\mathit{}$, where $t\mathit{}$ (time) is in seconds. When will the velocity be between $32 \;m/sec$ and $64 \;m/sec$? $\left(1, \dfrac{3}{2}\right)$ $\left(\dfrac{1}{2}, 1\right)$ $\left(\dfrac{1}{2},\dfrac{3}{2}\right)$ $\left(1, 3\right)$ Quantitative Aptitude gate2013-ae quantitative-aptitude speed-time-distance + – Akash Kanase asked Feb 15, 2016 edited Dec 7, 2017 by pavan singh Akash Kanase 2.9k views answer comment Share Follow See 1 comment See all 1 1 comment reply srestha commented Feb 22, 2016 reply Follow Share cannot understand symbols 1 votes 1 votes Please log in or register to add a comment.
Best answer 7 votes 7 votes Given $32< 80 - 32 t\mathit{} < 64$ $-48 < - 32 t\mathit{} < - 16$ $48 > 32 t\mathit{} > 16$ $\dfrac{3}{2} > t\mathit{} > \dfrac{1}{2}$ so ans is option c) $\left(\dfrac{1}{2} ,\dfrac{3}{2}\right)$ minal answered May 14, 2016 edited Dec 7, 2017 by pavan singh minal comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Speed of object gets slower when it moves upwards so first object attain speed 64 then 32 now, 80-32t=64 t=1/2 then 80-32t=32 t=3/2 Answer is C (1/2,3/2) Peeyush Pandey answered Dec 18, 2018 Peeyush Pandey comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes $\text{given relation is } v=80-32t ~$ $\implies 32t=80-v$ $\implies t=\frac{80-v}{32}$ $\text{now substituting v=32 we get } t=\frac32 $ $\text{and substituting v=64 we get } t=\frac12 $ $the~range~is~\left(\frac12,\frac32\right)$ subbus answered Jun 19, 2020 subbus comment Share Follow See all 0 reply Please log in or register to add a comment.