Answer:
(A) Constant acceleration will be [tex]a=1.479m/sec^2[/tex]
(B) Velocity of the car when it passes 0.14 km is 20.349 m/sec
Explanation:
It is given that driver starts from rest so initial velocity of the driver u = 0 m/sec
Distance traveled s = 0.14 km = 140 m
Time taken to cross 0.14 km is 8 sec
So time taken t = 8 sec
(A) From second equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]
So [tex]140=0\times 8+\frac{1}{2}\times a\times 8^2[/tex]
[tex]a=1.479m/sec^2[/tex]
(B) From third equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]v^2=0^2+2\times 1.479\times 140[/tex]
[tex]v^2=414.12[/tex]
v = 20.349 m/sec
So speed of the car when it passed 0.14 km is 20.349 m/sec