question_answer

The brakes applied to a car produce an acceleration of 6 m/s2 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.

Answer:

                                                                  Final velocity, v = 0, (\[\because \]Car stops after applying the brakes) \[a=-6\,m/{{s}^{2}},t=2s\] From first equation of y = u + at or\[u=\upsilon -at=0-(-6)(2)=12\,\text{m/s}\]             \[s=ut+\frac{1}{2}a{{t}^{2}}=(12)(2)+\frac{1}{2}(-6){{(2)}^{2}}\] = 12