Asked by Andre'
An object moves with constant acceleration 4.95 m/s2 and over a time interval reaches a final velocity of 13.2 m/s.
(a) If its initial velocity is 6.6 m/s, what is its displacement during the time interval?
(a) If its initial velocity is 6.6 m/s, what is its displacement during the time interval?
Answers
Answered by
Steve
v = at
13.2 = 4.95t
t = 8/3
s = vt + 1/2 at^2
s = 6.6*8/3 + 1/2 * 4.95 * (8/3)^2
s = 35.2
13.2 = 4.95t
t = 8/3
s = vt + 1/2 at^2
s = 6.6*8/3 + 1/2 * 4.95 * (8/3)^2
s = 35.2
Answered by
Willow Newman
a) 13.2
13.2^2=6.6^2+2(4.95)(ΔX)
174.24= 43.56+9.9ΔX
130.68=9.9ΔX
ΔX=13.2
13.2^2=6.6^2+2(4.95)(ΔX)
174.24= 43.56+9.9ΔX
130.68=9.9ΔX
ΔX=13.2
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