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The brakes applied to a car produce an acceleration of 6 ms2 in the opposite direction to the motion. If the car takes 2 \(s\) to stop after the application of brakes, calculate the distance it travels during this time.
 
Solution:
 
Time (\(t\)) \(=\) 2 \(s\)
 
Acceleration (\(a\)) \(=\) \(-\) 6 ms2 (Opposite direction to the motion)
 
Final Velocity (\(v\)) \(=\) \(0\) \(ms^{-1}\)
 
From the Equation of motion - I, We know that,
 
i=i+ii
 
Substituting the given values of \(v\), \(a\) and \(t\) in the above equation, we get
 
\(Initial\ velocity\ (u)\ =\)  \(ms^{-1}\)
 
From the Equation of motion - II, We know that,
 
i=ii+ii×iii
 
Substituting the given values of \(u\), \(a\) and \(t\) in the above equation, we get
 
Distance travelled \(=\)  \(m\).