Solution
According to given question car starts from rest and then comes to a stop by same acceleration and deacceleration and also time taken and distances travelled are same.
Now consider dt is time when car moves uniformly so acceleration/deacceleration time is given by as "\\frac{(t-dt) }{2}" .
So,
"<v>\\tau=2\\frac{w(\\tau-dt) ^2}{8}+\\frac{w(\\tau-dt).dt}{2}"
"(dt) ^2=\\tau^2-\\frac{4<v>\\tau}{w}"
"(dt) =\\tau\\sqrt{1-\\frac{4<v>}{w\\tau}}"
Now putting value of <v>=72km/h,
"\\tau=25s," "w=5m\/s^2"
Therefore car moves uniformly by time
"dt=15sec."
Comments
Leave a comment