Question

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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