A particle starts to move from rest in a straight line with an acceleration which increases with constant time rate from 1 ft./sec^{2} to 4 ft./sec^{2} in one second. prove that the particle will move a distance of 1 ft. in this second.

The particle's acceleration a increases at a constant rate, so that its average rate of change is equal to its instantaneous rate of change, which is equal to

(4 ft/s² - 1 ft/s²) / (1 s) = 3 ft/s³

The particles starts with acceleration a (0) = 1 ft/s², so we can determine its acceleration at time t by the fundamental theorem of calculus:

a(t) = a (0) + ∫₀ᵗ (3 ft/s³) du

a(t) = 1 ft/s² + (3 ft/s³) t

It also starts from rest, so with initial velocity v (0) = 0. Integrating again gives us the velocity function,

v(t) = v (0) + ∫₀ᵗ a(u) du

v(t) = ∫₀ᵗ (1 ft/s² + (3 ft/s³) u) du

v(t) = (1 ft/s²) t + 1/2 (3 ft/s³) t ²

Taking the particle's initial position to be x (0) = 0, compute the integral again to determine the distance it travels in 1 s:

x(t) = x (0) + ∫₀ᵗ v(u) du

x(t) = ∫₀ᵗ ((1 ft/s²) u + 1/2 (3 ft/s³) u ²) du

x(t) = 1/2 (1 ft/s²) t ² + 1/6 (3 ft/s³) t ³

→   x (1 s) = 1/2 (1 ft/s²) (1 s)² + 1/6 (3 ft/s³) (1 s)³ = 1 ft

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.