Question

An object with an initial velocity V0

speeds up with an acceleration a, travelling a distance L1

, then it slows

down with a deceleration a, and stops after travelling an additional distance L2

. If L2/L1=k, then what is themaximum velocity of the object during its travel?

(a${k-1/k+1} v0$

(b)$\sqrt{k/k-1} v0$

(c)${k/k-1} v0$

(d)$\sqrt{k+1/k} v0$

Answer 1

L2=kL1

first case

initial velocity=V0

during the acceleration of a,,

distance travelled=L1

Using third law of uniformly accelerated motion,we get

V^2=V0^2+2×a×L1

(Vmax)^2-Vo^2=2aL1.....i). (final velocity during acceleration will be maximum velocity)

now,,in second case(during deceleration)

S=L2

acceration=-a

initial velocity=final velocity , during acceleration i.e max. velocitty

also,V=0

so

0^2=(Vmax) ^2-2a×L2

Vmax^2=2aL2.....ii)

now,divide the eqation second from I)

(Vmax)^2-(V0)^2/(Vmax)^2=L1/L2

1-(V0)^2/(Vmax)^2=k...(L1/L2=K)

(1-k)/(V0)^2=1/(Vmax)^2

(Vmax)=Vo/(1-k)^1/2

{hope it helps}