An object travels with a speed 126km/h is bought to rest after travelling 80m. Find the acceleration and the time taken to stop the object.

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Solution

Here,
Initial speed, u = 126 km/h = 126 Ã— (5/18) m/s = 35 m/s
Distance, s = 80 m
Now,
v² = u² + 2as
=> 0 = 35² + 2a(80)
=> a = -7.7 m/s²
The negative sign indicates retardation.
Again,
v = u + at
=> 0 = 35 - 7.7t
=> t = 4.6 s
This is the time required to stop.

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An object travels with a speed 126km/h is bought to rest after travelling 80m. Find the acceleration and the time taken to stop the object.

Here,Initial speed, u = 126 km/h = 126 Ã— (5/18) m/s = 35 m/sDistance, s = 80 mNow,v2 = u2 + 2as=> 0 = 352 + 2a(80)=> a = -7.7 m/s2The negative sign indicates retardation.Again,v = u + at=> 0 = 35 - 7.7t=> t = 4.6 sThis is the time required to stop.

Here,
Initial speed, u = 126 km/h = 126 Ã— (5/18) m/s = 35 m/s
Distance, s = 80 m
Now,
v² = u² + 2as
=> 0 = 35² + 2a(80)
=> a = -7.7 m/s²
The negative sign indicates retardation.
Again,
v = u + at
=> 0 = 35 - 7.7t
=> t = 4.6 s
This is the time required to stop.