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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A car travels east at a speed of $30m/s^{-1}$ . It turns left through an angle of $90^{\circ}$ and continues to travel with the same speed. What is the change in its velocity ?

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If the initial and final velocity of the car are $\overrightarrow{V_1}$ and $\overrightarrow{V_2}$ respectively and $\Delta \overrightarrow{V}$ is the change to its velocity, then
$\overrightarrow{V_2} =\overrightarrow{V_1}+\Delta \overrightarrow{V}$
$\Delta \overrightarrow{V} =\overrightarrow{V_2}-\overrightarrow{V_1}=\overrightarrow{V_2}+(-\overrightarrow{V_1})$
To find $\Delta \overrightarrow{V}$ graphically , We must first draw a vector $\overrightarrow{AB}$ it represent $\overrightarrow{V_2}$ and then add a vector $\overrightarrow{BC}$ representing $-\overrightarrow{V_1}$
The vector sum of $\overrightarrow{V_2}$ and $-\overrightarrow{V_1}$ is the vector $\Delta \overrightarrow{V}$ which represented in magnitude and direction by $\overrightarrow{AC}$
As $|\overrightarrow{V_1}| =|\overrightarrow{V_2}|=30 ms^{-1}$ therefore $|\Delta \overrightarrow{V}| = 30 \sqrt 2 ms^{-1} $ and it is directed along north -west.
answered Jul 15, 2014 by meena.p
 

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