Question

A driver of a car travelling at 54 km per hour apply the brakes and accelerates uniformly in the opposite direction the car stop in 5 second another driver of another car travelling at 36 km per hour apply is break slowly and stop in 10 seconds find the distance travelled by the two car before stopping.

Answer 1

Car 1

Given :-

Initial velocity of Car 1= 54km/h or 15m/s

Final velocity= 0 [ start from rest]

Time (t) = 5 seconds

To find :-

Distance covered by the car

• First find the acceleration of car 1

$\boxed{\sf{\dag\ \ v= u+at}}$

$\longrightarrow\sf 0= 15+a \times 5\\ \\ \longrightarrow\sf -15= 5a\\ \\ \longrightarrow\sf \cancel{\dfrac{-15}{5}}= a\\ \\ \longrightarrow\sf -3= a$

$\underline{\bigstar{\sf{ Acceleration\ of \ Car \ 1 = {\boxed{\sf\ -3m/s^2}}}}}$

• Then finally find the distance covered by car 1
• By using 3rd equation of motion !

$\boxed{\sf{\dag\ \ v^2=u^2+2as}}$

$\longrightarrow\sf (0)^2= (15)^2+2\times (-3)\times s\\ \\ \longrightarrow\sf 0= 225-6\times s\\ \\ \longrightarrow\sf -225= -6s\\ \\ \longrightarrow\sf \cancel{\dfrac{-225}{-6}}= s\\ \\ \longrightarrow\sf s= 37.5m$

$\underline{\boxed{\sf{\ \ Distance \ covered \ by \ car_1= 37.5m}}}$

• Now the same process in finding the distance of Car2

Given :-

Initial velocity (u) =36 km/h or 10m/s

Final velocity (v)= 0 [ started from rest]

Time (t) = 10 seconds

To find :-

Distance covered by car2

• Find acceleration

• By 1st equation of motion v= u+at

$\longrightarrow\sf 0= 10+a \times 10\\ \\ \longrightarrow\sf -10= 10a\\ \\ \longrightarrow\sf \cancel{\dfrac{-10}{10}}= a\\ \\ \longrightarrow\sf -1= a$

$\underline{\bigstar{\sf{ Acceleration\ of \ Car \ 2 = {\boxed{-1m/s^2}}}}}$

Then finally find the distance covered by car 2

By using 3rd equation of motion !

$\boxed{\sf{\dag\ \ v^2=u^2+2as}}$

$\longrightarrow\sf (0)^2= (10)^2+2\times (-1)\times s\\ \\ \longrightarrow\sf 0= 100-2\times s\\ \\ \longrightarrow\sf -100= -2s\\ \\ \longrightarrow\sf \cancel{\dfrac{-100}{-2}}= s\\ \\ \longrightarrow\sf s= 50m$

$\underline{\boxed{\sf{\ \ Distance \ covered \ by \ car_2= 50m}}}$

Answer 2

GIVEN :-

• A driver of a car travelling at 54 km/h apply the brakes and accelerates uniformly in the opposite direction the car stop in 5 seconds.

• Another driver of another car travelling at 36 km/h apply is break slowly and stop in 10 seconds.

TO FIND :-

distance travelled by the two car before stopping.

FORMULA TO BE USED :-

• v = u + at

• v² - u² = 2as

SOLTUION :-

For car A,

Initial velocity(u) = 54 km/h

⇒u = 54 × 5/18

⇒u = 15 m/s

Time(t) = 5 s

Final velocity(v) = 0 m/s          

(As the car stops after 5 s)

Now,

$\sf{v=u+at}\\\\\sf{\implies 0=15+(a\times5)}\\\\\sf{\implies 5a=-15}\\\\\sf{\implies a=-3\ m/s^2}$

$\rule{130}{2}$

$\sf{v^2-u^2=2as}$

$\sf{\implies (0)^2-(15)^2=2\times-3\times s}\\\\\sf{\implies -225=-6s}\\\\\sf{\implies s=\dfrac{-225}{-6} }\\\\\boxed{\sf{\implies s=37.5\ m}}$

Hence, car A travels 37.5 m before coming to rest.

$\underline{\rule{200}{2}}$

For car B,

Initial velocity(u) = 36 km/h

⇒u = 35 × 5/18

⇒u = 10 m/s

Time(t) = 10 s

Final velocity(v) = 0 m/s

(As the car stops after 10 s)

Now,

$\sf{v=u+at}$

$\sf{\implies 0=10+(a\times10)}\\\\\sf{\implies -10=10a}\\\\\sf{\implies a=\dfrac{-10}{10}}\\\\\sf{\implies a=-1\ m/s^2}$

$\rule{130}{2}$

$\sf{v^2-u^2=2as}$

$\sf{\implies (0)^2-(10)^2=2\times-1\times s}\\\\\sf{\implies -100=-2s}\\\\\sf{\implies s=\dfrac{-100}{-2} }\\\\\boxed{\sf{\implies s=50\ m}}$

Hence, car B travels 50 m before coming to rest.