A particle moves along a straight line such that its acceleration is a = (4t^2-2) m/s, where t is in seconds. When t = 0, the particle is located 2 m to the left of the origin, and when t = 2, it is 20 m to the left of the origin. Determine the position of the particle when t=4s.

Answer with Explanations:

We are given:

a (t) = 4*t^2-2 ... (1)

where t = time in seconds, and a (t) = acceleration as a function of time.

and

x (0) = -2 ... (2)

x (2) = - 20 ... (3)

where x (t) = distance travelled as a function of time.

Need to find x (4).

Solution:

From (1), we express x (t) by integrating, twice.

velocity = v (t) = integral of (1) with respect to t

v (t) = 4t^3/3 - 2t + k1 ... (4)

where k1 is a constant, to be determined.

Integrate (4) to find the displacement x (t) = integral of (4).

x (t) = integral of v (t) with respect to t

= (t^4) / 3 - t^2 + (k1) t + k2 ... (5) where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x (0) = 0 - 0 + 0 + k2 = - 2 = > k2 = 2 ... (6)

substitute (6) in (3)

x (2) = (2^4) / 3 - (2^2) + k1 (2) - 2 = - 20

16/3 - 4 + 2k1 - 2 = - 20

2k1 = - 20-16/3 + 4 + 2 = - 58/3

=>

k1 = - 29/3 ... (7)

Thus substituting (6) and (7) in (5), we get

x (t) = (t^4) / 3 - t^2 - 29t/3 + 2 ... (8)

which, by putting t=4 in (8)

x (4) = (4^4) / 3 - (4^2 - 29*4/3 + 2

= 86/3, or

= 28 2/3, or

= 28.67 (to two places of decimal)