Question

13
A person moving on a motor cycle in a ground
takes a turn through 60° on his left after every
50 m. The magnitude of displacement suffered Q-
by him after 9th turn is
​

Answer 1

Answer:

50√2

Explanation:

at D the person will turn for the 9th time

displacement =(AD)^2=(AC)^2+ (DC)^2

=AD=50√2 m

Answer 2

The magnitude of the displacement suffered by the person moving a motor cycle is 100 m

Explanation:

On taking the turn of 60° every 50 m, the motorcyclist will complete a full hexagon after 6 turns and return to point A as shown in figure

On 7th , 8th and 9th turn he will go from A to G, then from G to H and then from H to I respectively

Initial starting point = A

Final ending point = I

Thus, the displacement = AI = AD

In ΔBCD

DC = BC = 50 m

∠BCD = 120°

∴ ∠CBD = ∠CDB = 30°

∴ ∠ABD = ∠ABC - ∠CBD = 120° - 30° = 90°

Thus ΔABD is a right angled triangle

In ΔBCD, by sine rule

$\frac{\sin 30^\circ}{CD}=\frac{\sin 120^\circ}{BD}$

$\implies \frac{1/2}{50}=\frac{\sqrt{3}/2}{BD}$

$\implies BD=50\sqrt{3}$

Therefore, in ΔABD

$AD^2=AB^2+BD^2$

$\implies AD^2=50^2+(50\sqrt{3})^2$

$\implies AD^2=50^2(1+3)$

$\implies AD^2=50^2\times 4$

$\implies AD=50\times 2=100$ m

Therefore, the magnitude of displacement is 100 m

Hope this answer is helpful.

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