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.
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to:
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to: