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

Step-by-step 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* 2.6* 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* (-1.5)* s_1`

`s_1=160.33 m`

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