Question

Part of a machine moves with a velocity given by the expression: 𝑣=7𝑡2+0.8𝑡 Where; v = velocity in mm/s t = time in seconds i. Calculate the displacement over the first 5 seconds of motion.

Answer 1

Given

$$v(t)=7t^2+0.8t$$

Then

$$r(t)=\int(7t^2+0.8t)dt$$

$$=\dfrac{7}{3}t^3+0.4t^2+ r_0$$

Displacement is

$$\Delta r(t)=\dfrac{7}{3}t^3+0.4t^2$$

$\Delta r(5)=\dfrac{7}{3}(5)^3+0.4(5)^2=\dfrac{905}{3}(mm)$

The displacement over the first 5 seconds of motion is $\dfrac{905}{3}$ mm or approximately $301.667$ mm.