Question

A block slides down an inclined plane, starting from rest and being pushed with a constant acceleration of 5.25 m/s2 over a distance of 18.0 cm. It then decelerates at a constant rate of 1.10 m/s2 (because of friction), until it again comes to rest. Find the total time the block is in motion.

Answer #1

As we know that

1st Part

givens:

vo = 0 m/s

a = 5.25 m/s^2

d = .18m

find:

time:

d = (vo)(t) + (1/2)(a)(t^2)

.18 = (0)(t) + (1/2)(5.25)(t^2)

.18 = 2.625t^2

t^2 = .068571428

t = 0.261861468 seconds

final velocity:

vf = vo + at

vf = 0 + 5.25(0.261861468)

vf = 1.374772708 m/s

2nd Part:

givens:

a = -1.10

vo = 1.374772708 (from final velocity of 1st part)

vf = 0

find:

time:

vf = vo + at

0 = 1.374772708 + (-1.10)(t)

1.10t = 1.374772708

t = 1.249793371

total time: 1.249793371 + .261861468 = 1.511654839
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