Question

a car accelerates from rest for a distance \[\alpha s \] , after which it decelerates for a distance \[\beta s\] and comes to rest in total time t..find the maximum velocity and initial acceleration?

Answer 1

v max =2s \[(\alpha + \beta)/t\] a1=\[2s(\alpha+\beta)^{2}/\alpha t ^{2}\]

Answer 2

this is the answer

Answer 3

apply s=ut+1/2a*t^2

for the first interval we have t=t1 & u=0 & s=alpha S hence
alpha s=0+1/2*a1*t1^2

also for this interval v=u+a t1
or v=0+a*t1

for the next interval we have
beta s=at1*(t-t1)+1/2*a2*(t-t1)^2
& 0=at1+a2*(t-t1)

these are the set of equations to solve? can u solve it? i am solving right now..

Answer 4

i will try to solve

Answer 5

did nt understand this beta s=at1*(t-t1)+1/2*a2*(t-t1)^2

Answer 6

yeah i reached the first reult..

Answer 7

ok plzzz go on

Answer 8

from this graph we have
area of 1st trinagle
\[\alpha s=1/2*t1*V_{\max}\]

area of 2nd triangle
\[\beta s=1/2*V_{\max}*(t-t1)\]

we solve this two equation
from ist we have
Vmax=2*alpha s/t1

from 2nd we have
Vmax=2*beta s/(t-t1)

we equate the above two equations to find the value of t1 as t1=alpha t/(alpha + beta)

put this in equation for Vmax in 1st equation we get Vmax as desired.