Question

15. When a car is stopped by applying brakes, it stops
after travelling a distance of 100 m. If speed of car
is halved and same retarding acceleration is applied
then it stops after travelling a distance of
[NCERT Pg. 50]
(1) 25 m
(2) 50 m
(3) 75 m
(4) 100 m​

Answer 1

Given:-

First Case

• Final Velocity of Car = 0m/s

• Distance Covered = 100m

Second Case :-

• Final Velocity = 0m/s

• Initial Velocity = $\sf{\dfrac{ u }{ 2 }}$

To Find:-

• If speed of car is halved and same retarding acceleration is applied then it stops after travelling a distance ?

Formulae used:-

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

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance

How To Solve ?

• Here, we don't know the value of "a" So, we will use the Third Equation of motion to find out the relation between "u" and "a"

• We will use the given value of 1st case to find the relation. then we will substitute the equation made by using them in the third equation of motion for 2nd case.

First Case:-

→ v² - u² = 2as

→ (0)² - (u)² = 2 × a × 100

→ -u² = 200a

→ a = -u²/200..........1

Therefore,

Second Case,

→ v² - u² = 2as

→ $\sf{v^2 - (\dfrac{u}{2})^2 = 2 * \dfrac{-u^2}{200} * s }$

→ $\sf{ 0^2 - \dfrac{u^2}{ 4 } = \dfrac{-u^{2}s}{100}}$

→ $\sf{ \dfrac{-u^2}{4} * 100 = -u^{2}s}$

→ $\sf{ -25u^2 = -u^{2}s}$

→ $\sf{ s = \dfrac{25\cancel{u^2}}{\cancel{u^2}}}$

→ $\sf{ s = 25m}$

Hence, The Care will cover 25m, Option 1 is correct.

Answer 2

Hope the picture is visible and you have understood the question!