A bus is moving with the initial velocity of 'u' m/s .After applying the breaks its retardation is 0.5 m/s2 and it stopped after 12s.Find the initial velocity (u) and the distancetrevelled by the bus after applying the breaks.
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Answer:
Okay Here is your answer
Explanation:
Let us write the given data first,
- Initial Velocity (u) = U m/s
- Retardation ( Deceleration ) a = -0.5 m/s^²
- Final Velocity (V) = 0 m/s {given that it is stopped}
- Time (t) = 12 sec
Solution:
To find U = ?
V = u+at
0 = U + (-0.5)× 12
0 = U - 6
U = 6m/s
To find Distance (s) travelled by bus,
v² - u² = 2 as
(0)² - (6)² = 2×(-0.5)×s
-36 = -1×s
S = 36 m
Hence the bus would have travelled 36m after applying the break
EXPLORE MORE:
❆ Use the newtons 3 laws of motion according to given quantities and find the answers . They are,
❆ v = u + at
❆ s = ut + 1/2 at²
❆ v² - u² = 2as
❆ Sn = u + (2n -1) used to find time in nth seconds
Hope it helps...
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