Question

a train is traveling with a velocity of 44m/sec. when break are applied it comes to rest after traveling a distance 121m. find the value of retardation produced by the breaks. how much time will the train take to come to rest.
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Answer 1

Given :-

☄ Initial velocity, u = 44m/s

☄ finial Velocity, v = 0m/s

☄ distance travelled, s = 121m

To find :-

the value of retardation produced by the breaks and the time taken to come to rest

Solution :-

using equation of motion .i.e.,

➠ v² = u² + 2as

➠ 0² = 44² + (2)(a)(121)

➠ 0 = 1936 + 242a

➠ -1936 = 242a

➠ a = -1936/242

➠ a = -8m/s²

|| -ve sign detects that it is retardation ||

thus, retardation producted by the break is -8m/s²

Now, using 1st equation of motion .i.e.,

➠ v = u + at

➠ 0 = 44 + (-8)(t)

➠ -44 = -8t

➠ t = 44/8

➠ t = 5.5sec

thus, the train will come to rest after 5.5sec .

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Answer 2

Answer:

Given :-

☄ Initial velocity, u = 44m/s

☄ finial Velocity, v = 0m/s

☄ distance travelled, s = 121m

To find :-

the value of retardation produced by the breaks and the time taken to come to rest

Solution :-

using equation of motion .i.e.,

➠ v² = u² + 2as

➠ 0² = 44² + (2)(a)(121)

➠ 0 = 1936 + 242a

➠ -1936 = 242a

➠ a = -1936/242

➠ a = -8m/s²

|| -ve sign detects that it is retardation ||

thus, retardation producted by the break is -8m/s²

Now, using 1st equation of motion .i.e.,

➠ v = u + at

➠ 0 = 44 + (-8)(t)

➠ -44 = -8t

➠ t = 44/8

➠ t = 5.5sec

thus, the train will come to rest after 5.5sec .