a body falls from rest it covers as much as distance in the last second of its motion as covered in the first three seconds the body has fallen for a time
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U = 0
Distance covered in first 3 s = 1/2a(3)(3)
Total time = t + 3
Distance covered in last second
=1/2 a [2(t+3) - 1]
= 1/2 a (2t + 5)
So 1/2 a (9) = 1/2 a (2t+5)
After solving, t = 2
So, Total time = t+3 = 5 seconds
Hope it helps...... ☺
Distance covered in first 3 s = 1/2a(3)(3)
Total time = t + 3
Distance covered in last second
=1/2 a [2(t+3) - 1]
= 1/2 a (2t + 5)
So 1/2 a (9) = 1/2 a (2t+5)
After solving, t = 2
So, Total time = t+3 = 5 seconds
Hope it helps...... ☺
rohirestle:
so total time is t+5 ha
no total time 5 sec hi hai
srry assumption as t+3
hmm
i couldnt understand this step 1/2 a [2(t+3) - 1]
i had applied the formula to calculate displacement at nth second
thats u say
which is [u + 1/2 a (2n-1)]
hey i got yar
hmm
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