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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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The velocity of a particle is given by $v=5-t^2 m/s.$ The average acceleration of the particle between 3 and 5 seconds will be

\[(a)\;2 m/s^2 \quad (b)\;-8 m/s^2 \quad (c)\;-14 m/s^2 \quad (d)\;9 m/s^2\]

Can you answer this question?
 
 

1 Answer

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At $t=3$
$v_1=5-(3)^2=-4$
at $t=5$
$v_2=5-(5)^2=-20$
Average acceleration$=\large\frac{v_2-v_1}{\Delta t}$
$\qquad=\large\frac{-20-(-4)}{2}$
$\qquad=\large\frac{-20+4}{2}$
$\qquad=-8 m/s^2$
Hence b is the correct answer. 

 

answered Jun 27, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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