A car moving with a speed of 12 m/s is subjected to brakes which produces a deceleration of 6 m/s2. The car takes 2 s to stop after the application of brakes. What is the distance covered by the car after the application of brakes?
12 m
24 m
36 m
48 m
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CDS-1 – 2023
Initial velocity, $u = 12$ m/s
Deceleration, $a = -6$ m/s² (negative because it’s deceleration)
Final velocity, $v = 0$ m/s (the car stops)
Time, $t = 2$ s
We need to find the distance covered, $s$.
We can use the equation of motion $s = ut + (1/2)at^2$.
Substituting the given values:
$s = (12 \, \text{m/s}) \times (2 \, \text{s}) + (1/2) \times (-6 \, \text{m/s}^2) \times (2 \, \text{s})^2$
$s = 24 \, \text{m} + (1/2) \times (-6) \times 4 \, \text{m}$
$s = 24 \, \text{m} – (3) \times 4 \, \text{m}$
$s = 24 \, \text{m} – 12 \, \text{m}$
$s = 12 \, \text{m}$
Alternatively, using $v^2 = u^2 + 2as$:
$0^2 = (12)^2 + 2 \times (-6) \times s$
$0 = 144 – 12s$
$12s = 144$
$s = 144 / 12 = 12$ m.
– Deceleration is negative acceleration.
1. $v = u + at$
2. $s = ut + (1/2)at^2$
3. $v^2 = u^2 + 2as$
4. $s = (u+v)t/2$
These equations are valid only when acceleration is constant.