Aparticle travels along a straight line with a velocity v = (12 – 3t2) m/s. when t = 1 s, the particle is located 10 m to the left of the origin. determine the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period.

The displacement from t = 0 to t = 10 s,  is -880 m

Distance is 912 m

Explanation:

.  .  . . . . . . . .  A

integrate above equation we get

from information given in the question  we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

we know that acceleration is given as

[FROM EQUATION A]

Acceleration at  t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

the distance the particle travels during this time period:

let v = 0,

t = 2 s

Distance

it was a fall down when it happens

explanation:

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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