Question

Two cars are travelling towards each other at a speed of \( 20 \, \text{m/s} \) each. When the cars are \( 300 \, \text{m} \) apart, both the drivers apply brakes and the cars retard at the rate of \( 2 \, \text{m/s}^2 \). The distance between them when they come to rest is:

• A. 200 m
• B. 50 m
• C. 100 m
• D. 25 m

Answer 1

The Correct Option is C

To solve this problem, we need to determine the distance between two cars when they come to rest after braking. Let's break it down step by step:

1. Both cars are traveling towards each other at a speed of \(20 \, \text{m/s}\).
2. The initial distance between the cars is \(300 \, \text{m}\).
3. The cars start applying brakes with a deceleration (negative acceleration) of \(2 \, \text{m/s}^2\).

We use the kinematic equation for constant acceleration:

\(v^2 = u^2 + 2as\)

• Here, \(v\) is the final velocity (0 m/s, as the cars come to rest),
• \(u\) is the initial velocity (\(20 \, \text{m/s}\)),
• \(a\) is the acceleration (\(-2 \, \text{m/s}^2\)),
• \(s\) is the distance covered after applying brakes until the car stops.

Apply the equation separately for each car:

1. For one car:
   • \(0 = (20)^2 + 2(-2) \times s_1\)
   • \(0 = 400 - 4s_1\)
   • Solving for \(s_1\) gives: \(s_1 = \frac{400}{4} = 100 \, \text{m}\)
2. Both cars travel \(100 \, \text{m}\) each before stopping.

The combined distance covered by both cars is \(100 \, \text{m} + 100 \, \text{m} = 200 \, \text{m}\). Since the initial distance between the cars was \(300 \, \text{m}\), the distance remaining between the two cars when they stop is:

\(300 \, \text{m} - 200 \, \text{m} = 100 \, \text{m}\)

Therefore, the distance between the cars when they come to rest is 100 m.

Answer 2

Let the distance between the cars when they come to rest be \( d \). Each car has an initial speed of 20 m/s. The deceleration \( a \) is \(-2 \, \text{m/s}^2\).

Using the equation of motion:

\( v^2 = u^2 + 2ad, \)

where \( v = 0 \) (final speed), \( u = 20 \, \text{m/s} \) (initial speed), and \( a = -2 \, \text{m/s}^2 \), we get:

\( 0 = 20^2 + 2(-2)d \quad \Rightarrow \quad 0 = 400 - 4d \quad \Rightarrow \quad d = 100 \, \text{m}. \)

Since the two cars are moving towards each other, the total distance covered by both cars is \( 100 + 100 = 200 \, \text{m} \), so the distance between them when they come to rest is 100 m.