Question

In a bike race, bike A takes 2 seconds
less than bike B at the finishing line and
passes the finishing line with a velocity v
more than the bike B. Assuming that the
bikes start from rest and travel with
constant accelerations 5 m/s2 and 4
m/s2 respectively, then vis equal to​

Answer 1

Answer:

Consider that A takes t

1

second, then according to the given problem, B will take (t

1

+t) seconds. Further let v

1

be the velocity of B at finishing point, then velocity of A will be (v

1

+v). Writing equations of motion for A and B

v

1

+v=a

1

t

1

....(i)

From equations (i) and (ii), we get

v=(a

1

−a

2

)t

1

−a

2

t .....(iii)

Total distance travelled by both the cars is equal

S

A

=S

B

⇒

2

1

a

1

t

1

2

=

2

1

a

2

(t

1

+t)

2

⇒t

1

=

a

1

−

a

2

a

2

t

Substituting this value of t

1

in equation (iii), we get

v=(

a

1

a

2

)t