Question

An object starts from rest from the point A to rest at point B on the same straight line at a distance d. It moves over the first part of the distance with an acceleration `alpha m//s^(2)` and for the remainder the retardation `beta m//s^(2)`. Find the time taken to complete the journey.

Answer 1

Method-I
Let C be the position where the speed of object is maximum `(v_(0))`. For A to C motion is uniformly accelerated and from C to B motion is retarded. We have
`d=AC+CB`
`rArr" "d=(v_(0)^(2))/(2alpha)+(v_(0)^(2))/(2beta)" "("use "v^(2)=u^(2)+2as" for AC and CB")`
`rArr" "v_(0)=sqrt((2alphabeta)/(alpha+beta)d)`
Total time taken

`t=t_(ArarrC)+t_(CrarrB)`
`rArr" "t=(v_(0))/(alpha)+(v_(0))/(beta)" (use v = u + at for AC and CB)"`
`rArr" "t=(v_(0)(alpha+beta))/(alpha beta)=sqrt((2alpha beta d)/((alpha+beta))).((alpha+beta)/(alpha beta))`
`rArr" "t=sqrt((2(alpha+beta))/(alpha beta)d)`
Method II Using the v-t graph
The v-t graph of the motion is shown. For the first part, the slope of v-t graph is equal to `alpha` and for the second part slope is equal to `-beta`. We have
Total distance travelled = Area under v-t graph
`rArr" "d=(1)/(2)v_(0)t" ...(i)"`
Also, `t=(v_(0))/(tan theta_(1))+(v_(0))/(tan theta_(2))`

`rArr" "t=v_(0)((1)/(alpha)+(1)/(alpha))`
`rArr" "v_(0)=(alpha beta t)/(alpha+beta)" ...(ii)"`
From (i) and (ii),
`t=sqrt((2(alpha+beta)d)/(alpha beta))`