Question 14 Marks
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
View full question & 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 $=\frac{\text { Distance travelled }}{\text { Total time }}$
Distance $=15 km =15 \times 1000=15000 m$
Time $=1 hr =1 \times 60 \times 60=3600 sec$.
$s=\frac{15000}{3600}$ or $s=\frac{15 km }{1 \text { hour }}=15 km /$ hour
$=4.16 m / sec$. or $15 km /$ hour
4. Velocity $=\frac{\text { Distance travelled }}{\text { Total time }}$
Displacement $=9 km =9 \times 1000=9000 m$
Time $=1 hr =1 \times 60 \times 60=3600 sec$
$V =\frac{9000}{3600}$ or $V =\frac{9 km }{1 \text { hour }}=9 km /$ hour
$=2.5 m / sec$. or $9 km / hour$
5. Yes, this velocity can be called as average velocity.
Actual distance $=15 km$
2. Displacement $=\overline{ AB }+\overline{ BD }+\overline{ DE }$ $=3+3+3$
Displacement $=9 km$
3. Speed $=\frac{\text { Distance travelled }}{\text { Total time }}$
Distance $=15 km =15 \times 1000=15000 m$
Time $=1 hr =1 \times 60 \times 60=3600 sec$.
$s=\frac{15000}{3600}$ or $s=\frac{15 km }{1 \text { hour }}=15 km /$ hour
$=4.16 m / sec$. or $15 km /$ hour
4. Velocity $=\frac{\text { Distance travelled }}{\text { Total time }}$
Displacement $=9 km =9 \times 1000=9000 m$
Time $=1 hr =1 \times 60 \times 60=3600 sec$
$V =\frac{9000}{3600}$ or $V =\frac{9 km }{1 \text { hour }}=9 km /$ hour
$=2.5 m / sec$. or $9 km / hour$
5. Yes, this velocity can be called as average velocity.