Question

A motor cycle starting from rest has an acceleration of 2.6 m s 2 When the motorcycle has traveled a distance of 120 m it slows down with an acceleration of 1.5 m s 2 until its velocity is 12 m s What is the total displacement of the motorcycle

Answer 1

Answer:280.33 m

Explanation:

Given

motorcycle starts from rest

initial acceleration$=2.6 m/s^2$

distance traveled =120 m

velocity gain during this time, let say v

$v^2-u^2=2as$

$v^2=2\times 2.6\times 120=24.97 \approx 25 m/s$

after it starts decelerating with 1.5 m/s^2 [/tex]

final velocity($v_f$)=12 m/s

$v_f=v+at$

12=25+(-1.5)t

t=8.667 s

distance traveled during this time

$v_f^2-v^2=2as_1$

$12^2-25^2=2\times (-1.5)\times s_1$

$s_1=160.33 m$

Therefore total displacement=$s+s_1=120+160.33=280.33 m$