Question

driver of a car travelling at 72 km/h applies the brakes and accelerates
uniformly in the opposite direction. The car stops in 5secs. Another driver
going at 36 km/h in another car applies his brakes slowly and stops in 10
secs. Which of the two cars travelled farther after the brakes were applied?

Answer 1

Answer:

Both are equal with 50m

Explanation:

Given :

1st car:

Initial velocity of car (u) = 72 km/h = 72 ×5/18 = 20 m/s

Time = 5 seconds

Final velocity = 0 m/s (As car finally comes to rest)

2nd car:

Initial velocity of car (u) = 36 km/h = 36×5/18 = 10 m/s

Time = 10 seconds

Final velocity =0 m/s (As car finally comes to rest)

Acceleration and distance of first car:

Using first equation of motion,

V=u+at

0=20+a×5

a = - 4 m/s^2

Now distance :

V^2-u^2=2as

0^2-20^2=2×-4×s

s= 50 metres

Acceleration and distance of second car:

Using first equation of motion,

V=u+at

0=10+a×10

a = - 1 m/s^2

Now distance :

V^2-u^2=2as

0^2-10^2=2×-1×s

s= 50 metres

Hence the distance covered by both the cars are equal to 50m

Answer 2

Given :-

Initial velocity of Car A (u)= 72km/h

convert in m/s

$\sf \cancel{72}\times \dfrac{5}{\cancel{18}}= 4\times 5= 20m/s\\ \\ \sf \bullet Speed \ of \ Car A= 20m/s\\ \\ \sf\bullet final \ velocity (v)=0\\ \\ \sf\bullet Time (t)= 5 seconds$

To find :-

The acceleration and distance of Car A

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

$\implies \sf 0= 20+a\times 5\\ \\ \implies \sf -20= 5a\\ \\ \implies \sf a=\cancel{\dfrac{-20}{5}}\\ \\ \implies \sf a= -4m/s^2$

Now find the distance covered by Car A

$\bigstar{\boxed{\sf{s=ut+\dfrac{1}{2}at^2}}}$

$\implies \sf s= 20\times 5+\dfrac{1}{\cancel{2}}\times \cancel{-4}\times (5)^2\\ \\ \implies \sf s=100+(-2\times 25)\\ \\ \implies \sf s= 100-50\\ \\ \implies \sf s=50m$

$\underline{\textsf{\textbf{\dag\ \ Distance \ covered \ by \ Car A = 50m}}}$

Now Car B

Initial velocity of Car B= 36km/h

convert in m/s

$\sf \cancel{36}\times \dfrac{5}{\cancel{18}}= 5\times 2= 10m/s\\ \\ \sf \sf\bullet Time (t) = 10seconds\\ \\ \sf\bullet final \ velocity (v)=0$

To find :-

The acceleration and distance of Car B

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

$\implies \sf 0= 10+a\times 10\\ \\ \implies \sf -10= 10a\\ \\ \implies \sf a=\cancel{\dfrac{-10}{10}}\\ \\ \implies \sf a= -1m/s^2$

Now find the distance of Car B

$\bigstar{\boxed{\sf{s=ut+\dfrac{1}{2}at^2}}}$

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

$\underline{\textsf{\textbf{\dag\ \ Distance\ covered \ by \ Car B= 50m}}}$

! Distance covered by both cars are equal to 50m