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Model Paper 8 — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 85 Marks
Question
On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ after every 500 m . Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
✓

Answer


Image
The path followed by the motorist is a regular hexagon with side 500 m , as shown in the given figure.
Let the motorist start from point P . The motorist takes the third turn at S .
Magnitude of displacement $= PS = PV + VS =500+500=1000 m(\because P V=Q R, V S=S R)$
Total path length, $d _1=P Q+Q R+R S=500+500+500=1500 m$
The motorist take the sixth turn at point P , which is the starting point
$\therefore$ Magnitude of displacement $=0$
Total path length, $d _2= PQ + QR + RS + ST + TU + UP$
$
d_2=500+500+500+500+500=3000 m
$
The motorist takes the eight turn at point R
$\therefore$ Magnitude of displacement $= PR$
$
\begin{aligned}
& P R=\sqrt{P Q^2+Q R^2+2(P Q) \cdot(Q R) \cos 60^{\circ}} \\
& P R=\sqrt{500^2+500^2+\left(2 \times 500 \times 500 \times \cos 60^{\circ}\right)} \\
& P R=\sqrt{250000+250000+\left(500000 \times \frac{1}{2}\right)} \\
& P R=866.03 m \\
& \beta=\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)=30^{\circ}
\end{aligned}
$
Therefore, the magnitude of displacement is 866.03 m at an angle of $30^{\circ}$ with PR.
Total path length $=$ Circumference of the hexagon $+ PQ + QR$
Total path length $=6 \times 500+500+500=4000 m$
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table
TurnMagnitude of displacement (m)Total path length (m)
Third10001500
Sixth03000
Eighth$866.03 ; 30^{\circ}$4000

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