# An Athelete completes one round of a circular track of radius 'R' in 40 seconds. What will be his displacement at the end of 2 minutes 20 seconds ?

one revolution completed in = 40 sec(s)

no. of revolutions in 2 min(s)20sec(s)= 140/40 = 3.5

i.e. three and a half revolutions.

then displacement is equal to diameter

• 8

Time taken to complete 1 revolution=40s

total time given=2min20s=(2*60)+20s=140s                           (1 minute=60s)

no. of revolutions=140/40=3.5

in 3 full rounds, displacement==0

but for the remaining .5 path left, the athlete can travel along the diameter of the circle as displacement is the2R shortest path taken

thus displacement=diameter of circle=

• 31

5.5
• -15
Displacement after 2 m 20 s i.e. in 140 s

Since,rotation in 40 s=1

Therefore, rotation in 1s=140

∴ rotation in 140s=140×140=3.5

Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m.

Therefore,

Distance covered in 2 m 20 s = 2198 m

And, displacement after 2 m 20 s = 200m

• -6
• 8 • 4
one revolution completed in = 40 sec(s)

no. of revolutions in 2 min(s)20sec(s)= 140/40 = 3.5

i.e. three and a half revolutions.?

then displacement is equal to diameter
• 4
2min 40 sec So 1 minute=60second So 2 minute =120 second 120+40=160 second Time taken to complete 1 round =40second 160÷40=4round 4(2πr)
• 2
Solution:-
1round=40s
Given time=2 min 20 s=2 ? 60 + 20 =140s

Number of rounds=140/40=3+1/2 round
Displacement in 3round = 0

Displacement in 1/2 round = 2R
• 1
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