Question

A car accelerates from rest with rate of alpha for some time and then deaccelerate with rate of Bita till to stop. if total time elappse 't' then find maximum velocity attends by the car and total distance cover

Answer 1

Let the car accelerate for t₁sec at a m/sec² and decelerate for t₂sec. at - b/ sec² 

At the end of t₁ sec. its velocity = 0 + at₁ m/sec. At the end of t₂ sec its velocity = 0 

Initial velocity at the beginning of deceleration is α t₁ 

V(t₂) = a t₁ - b t₂ = 0 Therefore t₂ = (a/ b ) t₁ ------------------------(1) 

But t = t₁ + t₂ ⇒ t = t₁ + t₂ = t₁ + (a/ b ) t₁ = t₁(1+ a/ b) 
∴ t₁ = t / (1+ a/ b) = {b / (a + b)}*t -------------------------------------- (2) 

Max. Velocity reached = a t₁ = a t / (1+ a/ b) = t / (1/ a + 1 / b) 

 = a*b*t/(a+b)

Distance covered =(a t₁² + b t₂²) /2 = ={a t₁² + b [(a/ b ) t₁)²]} /2 (substitute for t₂ from (1) 

Simplifying this = ½ * (a/ b) * (a + b)* t₁²

Now substitute for t₁ from (2) to get 
Distance covered = ½ *[a*b / (a +b)]*t²