Q: Obtain a relation for the distance travelled by am object moving with uniform acceleration in the interval between 4th and 5th sec.
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Answer:
distance travel between 4th and 5th sec
= S in 5th sec - S in 4th sec
= U + 1/2a{ 2 (5) - 1 } - { U + 1/2a ( 2×4 -1 )}
= 1/2a×9 - 1/2a×7
=a
Answered by
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Answer:
From second equation of motion, Distance travelled in t see Distance travelled in 4 s (distance travelled in 4th sec) Again, distance travelled in 5 s ( distance travelled in 5th sec) So, distance travelled in the interval between 4th and 5th second. So, the relation will be [u + 9/2a).
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