Question

A sliding mechanism in a machine moves with a velocity given by the expression:
v=7t+0.8t mm/s
calculate the displacement that occurs between t=2 and t=4 seconds

Answer 1

Answer on Question #71926-Physics-Mechanics-Relativity

A sliding mechanism in a machine moves with a velocity given by the expression:

$$v = 7t + 0.8t^2 \text{ mm/s}$$

Calculate the displacement that occurs between $t=2$ and $t=4$ seconds

Solution

The displacement that occurs between $t=2$ and $t=4$ seconds is

$$d = \int_{2}^{4} v(t) dt$$ $$d = \int_{2}^{4} (7t + 0.8t^2) dt = \left(\frac{7t^2}{2} + \frac{0.8t^3}{3}\right)_{2}^{4} = \left(7\frac{4^2}{2} + 0.8\frac{4^3}{3}\right) - \left(7\frac{2^2}{2} + 0.8\frac{2^3}{3}\right) = 57 \text{ mm} = 0.057 \text{ m}.$$

Answer: 0.057 m.

Answer provided by https://www.AssignmentExpert.com