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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A marble rolls across a flat surface. Its initial speed is 41 m/s but it comes to rest after 10 seconds . Find the acceleration of the marble and the distance traveled by the marble as it comes to rest.

$\begin{array}{1 1} -0.1 \;m/s^2,+5m \\ -0.2 \;m/s^2,+6m \\ 0.1 \;m/s^2,-5m \\-0.7 \;m/s^2,+5m\end{array} $

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We can use the first of our kinematic equation to solve for acceleration in terms of initial velocity, final velocity and time, all of which we know . Assuming the velocity to be positive the acceleration is :
$\large\frac{\overrightarrow{v} - \overrightarrow{v_0}}{t} $$=\overrightarrow{a} = \large\frac{0 m/s -1 m/s}{10 s}$$=-0.1 m/s^2$
From the second kinematic equation , which yields displacement fgrom initial velocity, acceleration and time , the distance traveled is:
$\overrightarrow{x}-\overrightarrow{x_0}=\overrightarrow{v}_0 t + \large\frac{1}{2}$$ \overrightarrow{a}t^2 =(1m/s) (10 s) + \large\frac{1}{2} $$(-0.1 m/s^2)(10 \;s)^2=+5 m $
answered Aug 19, 2014 by meena.p
Why you take velocity 1m/s inquestion it is given41m/s

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