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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 $64 \;m/sec$?

1.  $\left(1, \dfrac{3}{2}\right)$
2.  $\left(\dfrac{1}{2}, 1\right)$
3.  $\left(\dfrac{1}{2},\dfrac{3}{2}\right)$
4.   $\left(1, 3\right)$

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cannot understand symbols

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) $\left(\dfrac{1}{2} ,\dfrac{3}{2}\right)$

by
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)
$\text{given relation is } v=80-32t ~$

$\implies 32t=80-v$

$\implies t=\frac{80-v}{32}$

$\text{now substituting v=32 we get } t=\frac32$

$\text{and substituting v=64 we get } t=\frac12$

$the~range~is~\left(\frac12,\frac32\right)$
by