Question

Two cars A and B are approaching each other on
straight road with speed 72 km/h and 90 km/h
respectively. They spot each other when they are
500 m apart. If their reaction time is 0.5 s. Find
the distance between them if they apply breaks
producing same retardation of 5 m/s2
(1) 397.5 m
(2) 102.5 m
(3) 375 m
(4) 125 m​

Answer 1

Answer:

The distance between the cars after coming to stop is 80.5 m

Explanation:

Given that,

Total distance = 150 m

The velocity of both cars

v1=15m/s

v2=16m/s

The acceleration of both cars using the breaks

a1= -3m/s^2

a2= -4m/s^2

Case (I)

v1^2-u^2=2a1s1....

The distance of first car is

Using equation of motion

.....(I)

Where,

v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the equation (I)

0-15^2=2*-3*s1

s1=15*15/g

s1=37.5m

Case (II)

v2^2-u^2=2a2s2

The distance of second car is

Using equation of motion

.....(I)

Where,

v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the equation (I)

0-16^2=2*-4*s2

s2=32m

The distance between the cars after coming to rest is

s=150-(37.5+32)

s=80.5 m

Hence, The distance between the cars after coming to stop is 80.5 m.