Question

A bus slows down uniformly from 75.0 km/h to 0 km/h in 21 s. How far does it travel before stopping?

Answer 1

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s =  20-5/6 m/s

The average speed of the bus while it slows down is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s .

Traveling at an average of 10-5/12 m/s for 21 seconds,
the bus covers

                         (10-5/12) x (21) =  218.75 meters .

                  

Answer 2

As per the question,the bus slows down uniformly from 75 km/h to 0 km/h.

Hence  the initial velocity of the bus[ u]=75 km/hr.

             the final velocity of the bus [v]=0 km/h

             the taken by the bus to stop [t]=21 s

we know that 1 km= 1000 m and 1 hour =3600 second.

Hence $75 km/h =75*\frac{1000 m}{3600s}$

                          =20.83 m/s

we are asked to calculate the stopping distance.

From  equation of kinematics we know that-

                                    v= u +at   where a is the acceleration of the particle.

we have v= 0 km/h  = 0 m/s and u= 75 km/h =20.83 m/s

              t = 21 s

  Putting these values in above equation  we get-

                                 0 = 20.83 +a×21

                                ⇒ a×21 = -20.83 m/s

                                ⇒a=  -[20.83]÷21

                                     $= -0.9919 m/s^2$    [ Here negative sign indicates that particle is decelerating ]

Again from the equation of kinematics we  know that-

                                 $s = ut +\frac{1}{2} at^2$

Here s is the distance traveled. Putting the above quantities we get-

                                      $s = 20.83 *21 -\frac{1}{2} 0.9919 *[21^2]$

                                            s = 218.7 metre.  [ans]

Hence the bus will stop after a distance of 218.7 m.