Question

The following diagram shows the motion of a person along a straight horizontal path which is divided into two stages: In Stage I (AB), the person starts from rest and accelerates by a constant rate of 0.4 m/s over a distance of 20m. In stage 2 (BC), the person moves at constant velocity v for 10s. The average velocity of this person during the whole trip is:

Answer 1

Given

AB stage

v_o = initial velocity

v_o = 0 m/s

a = 0.40 m/s2

d = 20m

BC stage

v = constant

t = 10s

Procedure

To calculate the average speed we need to calculate the total distance and total travel time. To do this, let us calculate the final velocity at point B

`\begin{gathered} v_B^2=v_o^2+2ax \\ v_B=√(2ax) \\ v_B=√(2*0.4*20) \\ v_B=4m/s \end{gathered}`

Therefore, we can calculate the time on path AB as follows

`\begin{gathered} v_B=at \\ t=(v_B)/(a) \\ t=(4)/(0.4) \\ t=10s \end{gathered}`