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A car traveling at 40 m/s must come to a stop in 120 meters What (constant)...
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We can use the Equations of motion here.
`V^2 = U^2+2aS`
`V = 0` since the car comes to rest
`U = 40m/s`
`a = ???`
`S = 120m`
`a = (V^2-U^2)/(2S)`
`a = (0-40^2)/(2xx120)`
`a = -6.667m/s^2`
Using V = U+at
`0 = 40-6.667xxt`
`t = 6`
So the constant deceleration is 6.667 and the time required to stop is 6s.
Posted by jeew-m on September 3, 2013 at 10:41 AM (Answer #1)
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