Question

An
automobile
travelling
with
a
uniform
speed
of
60
km/h
on


applying
the
brakes
comes
to
a
stop
at
a
distance
of
20
m
with


uniform
retardation.
If
the
car
doubles
its
initial
speed,
what
will
be


the
new
stopping
distance?​

Answer 1

Answer:

80 m

is your answer I can't explain because explanation is more than 5000 word ok but 80 m is your answer I am also in your class

Answer 2

Given:

initial speed (u1)= 6km/h=6×(5/18)=(5/3)m/s

initial stopping distance=20m

speed after being doubled(u2)=(2×5/3)m/s=(10/3) m/s

find the stopping distance when speed is doubled?

Solution:

We know v²=u²+2as

• v after stopping is 0 (v is final velocity)

⇒0²=(5/3)²+2a×20

⇒0=25/9 + 40a

⇒40a=-25/9

⇒a=-25/(9×40)

⇒-a=5/72 ;   -a= retardation

Now lets  calculate distance taken to stop after double the speed

0²=(10/3)²+2(-5/72)(s)

100/9=10/72  (s)

s=(100/9)×(72/10)

=80m

s=80m