Answer :
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 .
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 [tex]75 km/h =75*\frac{1000 m}{3600s}[/tex]
=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
[tex]= -0.9919 m/s^2[/tex] [ Here negative sign indicates that particle is decelerating ]
Again from the equation of kinematics we know that-
[tex]s = ut +\frac{1}{2} at^2[/tex]
Here s is the distance traveled. Putting the above quantities we get-
[tex]s = 20.83 *21 -\frac{1}{2} 0.9919 *[21^2][/tex]
s = 218.7 metre. [ans]
Hence the bus will stop after a distance of 218.7 m.