Answer:
a = 0.0040 m/s², v = 14.4 m/s.
Step-by-step explanation:
Given that,
The distance between Kathmandu and Khanikhola, d = 26 km = 26000 m
Time, t = 1 hour = 3600 seconds
Let a is the acceleration of the bus. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Where
u is the initial speed of the bus, u = 0
So,
[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 26000}{(3600)^2}\\\\a=0.0040\ m/s^2[/tex]
Now using first equation of motion.
Final velocity, v = u +at
So,
v = 0+0.0040(3600)
v = 14.4 m/s
Hence, this is the required solution.
