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Calculate the acceleretion and the time to stop of 30m/s and 100m distance?
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Calculate the acceleretion and the time to stop of 30m/s and 100m dist...
Given= u=30m/s
v=0 m/s (the object stops)
s= 100m
Now, to find acceleration third equation of motion will be used. v2 = u2 + 2aS
Putting the values, we get
(0)^2 = (30)^2 +2.a.100
》 0 = 900 + 200a
》- 900 = 200a
》 -900÷200 = a
》 a = - 4.5 m/s^2
Therefore, acceleration (retardation) = -4.5/s^2

Now, to find time first equation of motion will be used
v=u+ at
Putting the values we get,
》0 = 30+ (-4.5)t
》-30= -4.5t
》30= 4.5t
》t= 30÷ 4.5
》t = 6.66 seconds

Therefore, acceleration is -4.5m/s^2
and time is 6.66 seconds.
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Calculate the acceleretion and the time to stop of 30m/s and 100m distance?
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