Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds. What
is the particle’s velocity when t = 6 s, and what is its position
when t = 11 s?
A particle travels along a straight line with a velocity
v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration
acceleration when
when t t == 44 s,s, the
the displacement
displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.
A particle travels along a straight line with a constant
acceleration. When s = 4 ft, v = 3 ft>s and when s = 10 ft,
v = 8 ft>s. Determine the velocity as a function of position.
SOLUTION
Velocity: To determine the constant acceleration ac, set s0 = 4 ft, v0 = 3 ft>s,
s = 10 ft and v = 8 ft>s and apply Eq. 12–6.
(:
+ ) v2 = v20 + 2ac (s - s0)
82 = 32 + 2ac (10 - 4)
ac = 4.583 ft>s2
Using the result ac = 4.583 ft>s2, the velocity function can be obtained by applying
Eq. 12–6.
(:
+ ) v2 = v20 + 2ac (s - s0)
v2 = 32 + 2(4.583) (s - 4)
A
v = A 29.17s - 27.7 ft>s Ans.
Ans:
v = ( 19.17s - 27.7 ) ft>s
4
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