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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Straight Line

A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of $350\;m/s$ The dart comes to rest in $0.0050\;s$ What is the acceleration of the dart ?

$\begin{array}{1 1} 7.0 \times 10^4 m/s^2 \\ 6.0 \times 10^3 m/s^2 \\ 2.0 \times 10^3 m/s^2 \\9.0 \times 10^3 m/s^2 \end{array} $

1 Answer

Answer : $ 7.0 \times 10^4 m/s^2$
$\overrightarrow{V_i}=350 \;m/s$
$V_f =0 \;m/s$
$ \Delta t= 0.0050\;s$
We may use the defining equation for acceleration .
$\overrightarrow{a_{av}}= \large\frac{\overrightarrow{V_f} -\overrightarrow {V_i}}{\Delta t}$
$\qquad= \large\frac{0 \Large\frac{m}{s} - 350 \large\frac{m}{s} }{0.0050\;s}$
$\qquad= -70000\;m/s^2 $
$\overrightarrow {a}_{av}=7.0 \times 10^4 m/s^2$
answered Aug 14, 2014 by meena.p
 

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