Question

A car accelerates from rest at a constant rate alpha for some time and reaches a maximum velocity.The n it decelerates at a constant rate beta to come rest.If the total time lapse d is t,then the maximum velocity acquired by the car will be

Answer 1

Answer:

Correct option is D)

Let maximum velocity=v

Now,v=0+αt

1

Similarly, 0=v−βt

2

From the above equations we get,

t

1

=

α

v

&t

2

=

β

v

t

1

+t

2

=t=

α

v

+

β

v

⇒v=

α+β

αβ

t

Answer 2

Please mark me as brainlist

Correct option is A)

Lets consider, v= maximum velocity,

When A car accelerates from rest at a constant rate α 

From 1st equation of motion,

v=u+at

v=0+αt1

v=αt1

t1=αv

After car decelerate at the rate β,

From 1st equation of motion,

v=u+at

0=v−βt2

v=βt2

t2=βv

Total time elapsed,

t=t1+t2

t=αv+βv

t=v(αβα+β)

v=(α+βαβ)t

From the above figure,

The total distance covered is equal to the area under the velocity-time graph

S=21×v×t

s=21(α+βαβ)t2