Question

A car starting from position of rest, moves with constant acceleration x. Then it moves with constant deceleration y and becomes stationary. If the total time elapsed during this is t, what would be the maximum velocity of the car?
(A) [xy/(x - y)] t
(B) [xy/(x + y)] t
(C) [x²y²/(x² + y²)] t
(D) [x²y²/(x² - y²)] t

Answer 1

Answer:

(xy/(x + y))T

Explanation:

Initial Velocity = 0

Let say T₁ time taken to reach Max velocity then

constant acceleration x

Max Velocity  = 0 + xT₁ = xT₁

Time Taken during deceleration = T - T₁

During decelration Initial Velocity = xT₁

final Veloicty = 0

0  = xT₁  - y(T - T₁)

xT₁ = y(T - T₁)

=> T₁(x + y) = yT

=> T₁ = yT/(x + y)

putting Value of T₁  in xT₁

Max Velocity = xyT/(x + y)

= (xy/(x + y))T

Answer 2

Answer:

Max velocity =Vmax= xyt/(x+y)

Explanation:

Given :

Case 1: during acceleration

Initial velocity=u= 0 /s

Acceleration = x

Time taken=t= t1 sec

From first equation of motion :

V=u+at

=0+xt1

Vmax=xt1

Case 2:

During Deceleration

U=xt1

Time =t

-t1

V=0 m/s  

Deceleration = -y

Time =t

0=xt1-y(t-t1)

0=xt1-yt+yt1

t1(x+y)=yt

t1= yt/x+y

Vmax= xyt/x+y