Observe the figure and answer the questions.Sachin and Sarneer started

1.	Observe the figure and answer the questions.Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?
Answer» 1. Actual distance = \(\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}\) = 3 + 4 + 5 + 3 Actual distance = 15 km 2. Displacement = \(\overline{AB}+\overline{BD}+\overline{DE}\) = 3 + 3 + 3 Displacement = 9 km 3. Speed = \(\frac{Distance\,travelled}{Total\,time}\) Distance = 15 km = 15 × 1000 = 15000 m Time = 1 hr = 1 × 60 × 60 = 3600 sec. s = \(\frac{15000}{3600}\) or s = \(\frac{15\,km}{1\,hour}\) = 15km/hour = 4.16 m/sec. or 15 km/hour 4. Velocity = \(\frac{Distance\,travelled}{Total\,time}\) Displacement = 9 km = 9 × 1000 = 9000 m Time = 1 hr = 1 × 60 × 60 = 3600 sec V = \(\frac{9000}{3600}\) or V = \(\frac{9\,km}{1\,hour}\) = 9 km/hour = 2.5 m/sec. or 9 km/hour 5. Yes, this velocity can be called as average velocity.

Observe the figure and answer the questions.Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?

Answer»

1. Actual distance = \(\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}\)

= 3 + 4 + 5 + 3

Actual distance = 15 km

2. Displacement = \(\overline{AB}+\overline{BD}+\overline{DE}\)

= 3 + 3 + 3

Displacement = 9 km

3. Speed = \(\frac{Distance\,travelled}{Total\,time}\)

Distance = 15 km = 15 × 1000 = 15000 m

Time = 1 hr = 1 × 60 × 60 = 3600 sec.

s = \(\frac{15000}{3600}\) or s = \(\frac{15\,km}{1\,hour}\) = 15km/hour

= 4.16 m/sec. or 15 km/hour

4. Velocity = \(\frac{Distance\,travelled}{Total\,time}\)

Displacement = 9 km = 9 × 1000 = 9000 m

Time = 1 hr = 1 × 60 × 60 = 3600 sec

V = \(\frac{9000}{3600}\) or V = \(\frac{9\,km}{1\,hour}\) = 9 km/hour

= 2.5 m/sec. or 9 km/hour

5. Yes, this velocity can be called as average velocity.