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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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3 Answers

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Best answer

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

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

$\implies 32t=80-v$

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

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

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

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

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