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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:

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 A class=

Sagot :

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

[tex]\begin{gathered} v_B^2=v_o^2+2ax \\ v_B=\sqrt{2ax} \\ v_B=\sqrt{2*0.4*20} \\ v_B=4m/s \end{gathered}[/tex]

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

[tex]\begin{gathered} v_B=at \\ t=\frac{v_B}{a} \\ t=\frac{4}{0.4} \\ t=10s \end{gathered}[/tex]

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