Question

1. The position of a particle which moves along the straight line is define by the relation
() = ^3 − 6^2 − 15 + 40 where and are expressed in meters and seconds respectively.
Note that the coefficients of t have dimensions accordingly.
(a) Determine when the velocity of the particle is zero.
(b)Calculate the position vector and distance travelled by the particle when the
acceleration is zero. Consider that at the starting point time = 0 sec.
(c)Does the particle move at constant velocity or constant acceleration? Justify
your answer.

Answer 1

Answer:

Explanation:

Answer:

solved

Explanation:

x = t³– 6t² – 15t + 40

(a) the time at which the velocity will be zero

x' = 3t²– 12t – 15 = 0

t²– 4t – 5 = 0

t = 5 sec.

(b) the position and distance traveled by the particle at that time

x = t³– 6t² – 15t + 40 = 5³-6.5² - 15.5+40 = - 60 feet

v = 3t²– 12t – 15 = 3.5²

(c) the acceleration of the particle at that time

v = 3t²– 12t – 15

a = 6t - 12 = 30 -12 = 18 f/s²

(d) the distance traveled by the particle from t = 4 s to t = 6 s.

x(t) = t³– 6t² – 15t + 40

x(4) = 4³– 6.4² – 15.4 + 40 = - 42 f

x(3) = 3³– 6.3² – 15.3+ 40 = -32 f

x(4) - x(3) = - 10 f