Question

A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5 s later. Find the distance that the car travels during braking

Answer 1

As per the question the initial speed of the car [ u] is 42 m/s.

The car applied its brake and comes to rest after 5.5 second.

The final velocity [v] of the car will be zero.

From the equation of kinematics we know that

                                         $v=u+at$ [ here a stands for acceleration]

                                         $0=42 +5.5a$

                                         $a =\frac{-42}{5.5} m/s^2$

                                          $a= -7.64 m/s^2$

Here a is taken negative as it the car is decelerating uniformly.

We are asked to calculate the stopping distance .

From equation of kinematics we know that

                                              $S=ut+\frac{1}{2} at^2$  [here S is the distance]

                                                      $= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m$

                                                       $=115.445 m$      [ans]