Question

$\orange\bigstar$$\underline{\boxed{\gray {\textbf{Question\::-}}}}$

Ramesh can bring a car at 30km/h to a halt by applying brakes with in a distance of 10m what distance would the same car cover before coming to rest if it's initial velocity be 60km/h​

Answer 1

Newton's motion equation:

v^2 = u^2 – 2as…….. where a: retardation (opposite of acceleration) and s: distance covered.

1st case:

u = 30 km/h = 25/3 m/s

v = 0,

s = 8 m

Therefore, 0^2 = (25/3)^2 - 2 × a × 8

This gives a = 4.34 m/s^2

2nd case:

u = 60 km/h = 50/3 m/s

v = 0

a = 4.34 m/s^2

Now, 0^2 = (50/3)^2 - 2 × 4.34 × s

This gives s = 32 m

Thus on applying brakes the car will stop after covering 32 m

Explanation:

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Answer 2

Given :

▪ Initial velocity of car A = 30kmph

▪ Stopping distance for car A = 10m

▪ Initial velocity of car B = 60kmph

To Find :

▪ Stopping distance for car B.

Concept :

↗ This question is completely based on the concept of stopping distance.

↗ Stopping distance is defined as the distance travelled by body before it is brought to rest by applying brakes. (constant retardation)

✴ Deceleration has said to be constant throughout the period of motion, we can easily apply third equation of kinematics to solve this question.

$\implies\bf\:v^2-u^2=2as\\ \\ \implies\sf\:(0)^2-u^2=2(-a)s\\ \\ \implies\sf\:u^2=2as\\ \\ \implies\underline{\boxed{\bf{\red{s=\dfrac{u^2}{2a}}}}}\:\orange{\bigstar}$

• v denotes final velocity
• u denotes initial velocity
• a denotes deceleration
• s denotes stopping distance

[Note : -ve sign of a shows retardation.]

Calculation :

$\dashrightarrow\sf\:\dfrac{S_A}{S_B}=\dfrac{(u_A)^2}{(u_B)^2}\rightarrow\:(\because\:a_A=a_B)\\ \\ \dashrightarrow\sf\:\dfrac{10}{S_B}=\dfrac{(30)^2}{(60)^2}\\ \\ \dashrightarrow\sf\:\dfrac{10}{S_B}=\dfrac{1}{4}\\ \\ \dashrightarrow\sf\:S_B=10\times 4\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{S_B=40m}}}}\:\gray{\bigstar}$

⚠ Stopping distance doesn't depend on the mass of moving body.