Distance travelled by a particle starting form rest and moving with an acceleration of 4/3ms-2 in the second is
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Distance travelled by a particle starting form rest and moving with an acceleration of 4/3ms-2 in the second is
Option 1 -
1
Option 2 -
2
Option 3 -
3
Option 4 -
4
-
1 Answer
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Correct Option - 2
Detailed Solution:Please find the solution below:
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[h] = ML2T-1
[E] = ML2T-2
[V] = ML2T-2C-1
[P] = MLT-1
According to question, we can write
10 =
Average speed
(d) Initial velocity
Final velocity
Change in velocity
Momentum gain is along
Force experienced is along
Force experienced is in North-East direction.
h = u²/2g, u = √2gh
Now, S = h/3
S = ut + ½at²
h/3 = √2ght - ½gt²
t² - 2√ (2h/g)t + 2h/3g = 0
Using quadratic formula for t:
t = ( 2√ (2h/g) ± √ (8h/g) - 4 (2h/3g) / 2
t = √ (2h/g) ± √ (2h/g - 2h/3g) = √ (2h/g) ± √ (4h/3g)
t? /t? = (√ (2h/g) - √ (4h/3g) / (√ (2h/g) + √ (4h/3g)
t? /t? = (√2 - √ (4/3) / (√2 + √ (4/3) = (√6 - 2)/ (√6 + 2)
(Note: There is a calculation error in the provided solution. Re-evaluating the physics.)
h/3 = (√2gh)t - ½gt²
(g/2)t² - (&
Air resistance resists the motion of an object. In this case, the net acceleration is lesser than 'g' and it shrinks as the speed increases. This makes the object to speed up more slowly. Ultimately, it reaches a constant terminal velocity which is lower for large-area ones and higher for heavy and streamlined ones.
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