Hi ppl....
PHYSICS... CLASS 9
I want ans for 1 , 2 ,3
Answers
Explanation:
HI there !!
1)
Displacement is the shortest distance possible between initial and final position.
As the athlete is in circular motion , displacement is zero. [ THis is because , the initial and final position is the same ]
Distance covered in 1 round = Circumference of the circle
Circumference of the circle = 2πr
Diameter = 200 m
Radius = 100 m
Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m
Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m
2min 20 sec = 140 sec
Distance covered in 140 sec = 15.71 × 140 = 2199.4 m
2)Velocity = dispacement / time
Speed = distance / time
a) when he jogs from A to B on a straight road,
displacement = distance = 300m
time = 2 minutes 30 seconds = 150 s
velocity = 300/150 = 2 m/s
speed = 300/150 = 2m/s
b)when he jogs from A to B and turns back to C,
displacement = 300-100 = 200m
distance = 300+100 = 400m
time = 3 minute 30 second = 210 s
velocity = 200/210 = 20/21 m/s
speed = 400/210 = 40/21 m/s
3)Let the distance traveled by Abdul from home to school = s km
time taken to reach the school=t1 sec
For Return journey , Abdul cover distance =s km
time=t2 sec
∴ Average speed for forward journey[home - school ] = Total distance/ Total time.
20 km/h =s/t1
∴t1=s/20 h ------eq(1)
Average speed for backward journey[school -home] = Total distance/ Total time.
30 km/h =s/t2
t2=s/30h -----eq(2)
Average distance for entire journey = Total distance/ Total time
=(s+s)/[s/20 +s/30]
=2s/s[1/20+1/30]
=2x20x30/50
=24 km/hr
Shortcut method : If equal distance are covered with speeds v1 and v2 then average distance=2v1v2/v1+v2
= 2x20x30/50
=24km/hr
∴ The average speed for Abdul's trip is 24km/h
hope this answer helpful u
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Explanation: