Question

A particle 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.

Answer 1

Answer:

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

Distance is 912 m

Explanation:

$v = (12 - 3t^2) m/s = ds/dt$.  .  . . . . . . . .  A

integrate above equation we get

$s = 12t - t^3 + C$

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

$s(t) = 12t - t^3 - 21$

we know that acceleration is given as

$a(t) = dv/dt = -6t$  

[FROM EQUATION A]

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

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

$s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m$

the distance the particle travels during this time period:

let v = 0,

$3t^2 = 12$

t = 2 s

Distance $= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m$