Question

please explain the question .the first answer that i will get from anyone i will mark brainleast .please do not be spam

Answer 1

Given:-

●  Two cars 1 and 2 are 80 m apart

● initial velocity of Car 1 , u₁ = 72 km/h

   so,   u₁ = 72 × 5 / 18 = 20 m/s

● initial velocity of Car 2 , u₂ = 60 km/h

  so, u₂ = 60 × 5 / 18 = 50/3   m/s

●On applying brakes, retardation  of Car 1 is equal to retardation of Car 2

  so, a₁ = a₂ = - 5 m/s²

To find:-

We have to determine whether the the Cars will avert the collision or not .

Formulae required:-

⇢  Third equation of motion

$\boxed{\bf{2\;a\;s=v^2-u^2}}$

where ,

a is the acceleration , s is the distance covered , v is the final velocity and u is the initial velocity.

Solution:-

Since , Breaks were applied to cars

hence , final velocity of Car 1 will be equal to final velocity of Car 2

so,

v₁ = v₂ = 0  m/s

__________________________________

▶Calculating distance covered by Car 1 after applying brakes

we have known values

a₁ = - 5 m/s²

u₁ = 20 m/s

v₁ = 0 m/s

Let,

distance covered by Car 1 after brakes = s₁

putting values in the third eqn of motion

$\implies\rm{2\;a_1\;s_1={v_1}^2-{u_1}^2}$

$\implies\rm{2\;(-5)\times s_1=(0)^2-(20)^2}$

$\implies\rm{-10\times s_1 = -400}$

$\implies\rm{\purple{s_1=40\:m}}$

▶Calculating distance covered by Car 2 after applying brakes

we have known values

a₂ = - 5 m/s²

u₂ = 50/3  m/s

v₂ = 0  m/s

Let,

distance covered by Car 2 after brakes = s₂

Putting values in third eqn of motion

$\implies\rm{2\;a_2\;s_2={v_2}^2-{u_2}^2}$

$\implies\rm{2(-5)\times s_2 = (0)^2 - {(\frac{50}{3})}^2}$

$\implies\rm{-10\times s_2=\frac{-2500}{9}}$

$\implies\rm{s_2=\frac{250}{9}}$

$\implies\rm{\purple{s_2=27.8\:m}}$

Now,

s₁ + s₂ < 80 m

It means The collision will be averted .