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4. An object moves with constant acceleration `4.00 m//s^2` and over a time interval reaches a final velocity of `12.0 m//s`.
(a) If its initial velocity is `6.00 m//s`, what is its displacement during the time interval?
(b) What is the distance it travels during this interval?
(c) If its initial velocity is `-6.00 m//s`, what is its displacement during the time interval?
(d) What is the total distance it travels during the interval in part (c) ?
`v_f^2 = v_i^2 +2a∆x`
(a)`13.5m`
(b)`13.5m`
(c)`13.5m`
(d)`22.5m`
Givens:
`a=4.00 m//s^2`
`v_f = 12.0 m//s`
(a)
Given:`v_i = 6.00 m//s`
`v_f^2 = v_i^2 +2a∆x`
`∆x = (v_f^2 - v_i^2)/(2a)`
`∆x = ((12.0 m//s)^2 - (6.00 m//s)^2)/(2(4.00 m//s^2)) = color(red)(13.5m)`
(b)
Since the acceleration and velocity of the object are in the same direction. distance equals displacement.
`d= color(red)(13.5m)`
(c)
Given:`v_i = -6.00 m//s`
`v_f^2 = v_i^2 +2a∆x`
`∆x = (v_f^2 - v_i^2)/(2a)`
`∆x = ((12.0 m//s)^2 - (-6.00 m//s)^2)/(2(4.00 m//s^2)) = color(red)(13.5m)`
(d)
Since the acceleration and velocity of the object are not in the same direction. distance equals displacement of `v= -6 m//s to v=0` plus displacement of `v=0 to v=12.0 m//s`.
Fist part:
`v_i = -6.00 m//s`
`v_f = 0`
`∆x_1 = ((0 m//s)^2 - (-6.00 m//s)^2)/(2(4.00 m//s^2)) = -4.50 m`
Second part:
`v_i = 0 m//s`
`v_f = 12.0 m//s`
`∆x_1 = ((12.0 m//s)^2 - (0 m//s)^2)/(2(4.00 m//s^2)) = 18.0 m`
distance:
`d = |∆x_1| + |∆x_2| = |-4.50 m| + |18.0 m| = color(red)(22.5m)`
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