Question

The position of a particle which moves along a straight line is defined by the relation x = t3
– 6t2
– 15t + 40, where x is expressed in feet and t in seconds. Determine (a) the time at which the velocity will be zero, (b) the position and distance traveled by the particle at that time, (c) the acceleration of the particle at that time, (d) the distance traveled by the particle from t = 4 s to t = 6 s.

Answer 1

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