On an open ground, a motorist follows a track that turns to his left by an angle of $60^?$ after every $500\,m$. Starting from a given turn, the displacement of the motorist at the third turn is
Updated On: Sep 8, 2023
$500\,m$
$500\sqrt{3}\,m$
$1000\,m$
$1000\sqrt{3}\,m$
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The Correct Option isC
Solution and Explanation
The path followed by the motorist is a regular hexagon $ABCDEF$ of side length $500\,m$ as shown in the figure.
Let the motorist starts from $A$ and takes the third turn at $D$.
Therefore, the magnitude of this displacement is
$AD = AG + GD$$ = 500 \,m + 500 \,m$$ = 1000\, m$
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
v = final velocity of the particle
u = initial velocity of the particle
s = displacement of the particle
a = acceleration of the particle
t = the time interval in which the particle is in consideration
