A toy cyclist completes one round of a square track of side 2 m in 40s . What will be the displacement at the end of 3 min?

Question

A toy cyclist completes one round of a square track of side 2 m in 40s . What will be the displacement at the end of 3 min? (A) 52 m (B) Zero (C) 16m (D)  2√2m

Tardigrade Admin · Accepted Answer

Displacement is distance from initial to final position. In 40 s-1 round In 3× 60=180 text s - 4(1/2) round Displacement for 4 rounds is zero. For (l/2) round, length of diagonal =2√2m .

Q. A toy cyclist completes one round of a square track of side $2$ m in $40 s$ . What will be the displacement at the end of $3$ min?

52 m

Zero

16m

$22 m$

$42 m$

Displacement is distance from initial to final position.
In $40 s - 1$ round In $3 \times 60 = 180 s - 4 \frac{1}{2}$ round Displacement for 4 rounds is zero.
For $\frac{l}{2}$ round, length of diagonal $= 22 m$ .

Q. A toy cyclist completes one round of a square track of side 2 m in 40s . What will be the displacement at the end of 3 min?

52 m

Zero

16m

22​m

42​m

Displacement is distance from initial to final position. In 40s−1 round In 3×60=180s−421​ round Displacement for 4 rounds is zero. For 2l​ round, length of diagonal =22​m .

Q. A toy cyclist completes one round of a square track of side $2$ m in $40 s$ . What will be the displacement at the end of $3$ min?

$22 m$

$42 m$

Displacement is distance from initial to final position.
In $40 s - 1$ round In $3 \times 60 = 180 s - 4 \frac{1}{2}$ round Displacement for 4 rounds is zero.
For $\frac{l}{2}$ round, length of diagonal $= 22 m$ .