A person swims 100m in the first 40 seconds 80m in the next 40 second and 40m in the last 20 seconds. What is the average speed?
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Let the initial velocity be ‘u’ and the constant acceleration be ‘a’.
Distance covered in first 2 seconds =2m
=> ut +½at²=2m
=>2u+2a=2m (t=2seconds)
=>u+a=1———(I)
Distance covered in last 4 seconds=2.2m
Total distance in 6 seconds =2.2+2=4.2m
=>ut+½at²=4.2m
=>6u+18a=4.2m (t=6 seconds)
=>u+3a=0.7-(ii)(divinding both sides by 7)
Now subtracting equation (ii) form (I),we get—
-2a=0.3
=>a= -0.15m/s²
Putting this value in equation (I), we have--
u=1.15m/s
Now , velocity at the end of 7th second = u+at = 1.15 + (-0.15*7)=1.15–1.05m/s = 0.1m/s
Hence, velocity at the end of 7th second is 0.1m/s or 10cm/s.
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Distance covered in first 2 seconds =2m
=> ut +½at²=2m
=>2u+2a=2m (t=2seconds)
=>u+a=1———(I)
Distance covered in last 4 seconds=2.2m
Total distance in 6 seconds =2.2+2=4.2m
=>ut+½at²=4.2m
=>6u+18a=4.2m (t=6 seconds)
=>u+3a=0.7-(ii)(divinding both sides by 7)
Now subtracting equation (ii) form (I),we get—
-2a=0.3
=>a= -0.15m/s²
Putting this value in equation (I), we have--
u=1.15m/s
Now , velocity at the end of 7th second = u+at = 1.15 + (-0.15*7)=1.15–1.05m/s = 0.1m/s
Hence, velocity at the end of 7th second is 0.1m/s or 10cm/s.
If you liked it then please upvote!!
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