Question

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 1

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)