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:
- 200 m
- 50 m
- 100 m
- 25 m
The Correct Option is C
Approach Solution - 1
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:
- Both cars are traveling towards each other at a speed of \(20 \, \text{m/s}\).
- The initial distance between the cars is \(300 \, \text{m}\).
- 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:
- For one car:
- \(0 = (20)^2 + 2(-2) \times s_1\)
- \(0 = 400 - 4s_1\)
- Solving for \(s_1\) gives: \(s_1 = \frac{400}{4} = 100 \, \text{m}\)
- 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.
Approach Solution -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.
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