Question

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, specify the displacement of the motorist at the 3rd, 6th and 8th turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case

Answer 1

The entire motion of the motorist can be shown using the below diagram,
Let A is the starting point of the motorist.
At the third turn:
At the third turn, his position is D.
Magnitude of Displacement = AD
so,  Magnitude of Displacement = AO + OD
= 500m + 500m
⇒ Magnitude of Displacement = 1000m

Total path length = AB + BC + CD
= 500m + 500m + 500m
So, Total path length = 1500m
At the sixth turn:
At the sixth turn, the position of the motorist is A.
So, the final position is the initial position i.e., there is no change in position.
Hence, magnitude of displacement = 0
Total path length = AB + BC + CD + DE + EF + FA
⇒ Total path length = 500m+500m+500m+500m+500m+500m
⇒ Total path length = 3000m
At the eighth turn:
At the eighth turn, the position of the motorist is C.
Magnitude of displacement = AC
⇒ Magnitude of displacement = $\sqrt{AB^2+BC^2+2AB.BCcos60^{\circ}}$
= $\sqrt{500^2+500^2+2(500)(500)cos60^{\circ}}$
=$\sqrt{500^2+500^+500^2}$
=$500\sqrt{3}$ ≈ 866m

Total path length = AB + BC
 = 500m + 500m
∴ Total path length = 1000m

Answer 2

For any arbitrary motion in space, which of the following relations are true: a. $V_{average}= (\frac{1}{2})[v(t_1)+v(t_2)]$ b. $V_{average}=\frac {[r(t_2)-r(t_1)]} {(t_2-t_1)}$ c. v (t) = v (0) + at d. r (t) = r (0) + v (0) t + (1/2) at² e. $a_{average}=\frac {[v(t_2)-v(t_1)]} {(t_2-t_1)}$ (The 'average' stands for average of the quantity over the time interval $t_1 to t_2$)