Answer:
(a) 508.37 m
(b) 47.53 s
(c) 21.165 m
(d) 19.365 m
Explanation:
initial velocity, u = 77 km/h = 21.39 m/s
acceleration, a = - 0.45 m/s^2
(a) final velocity, v = 0
Let the distance traveled is s.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=21.39^{2}-2 \times 0.45 \times s[/tex]
s = 508.37 m
(b) Let t be the time taken to stop.
Use first equation of motion
v = u + at
0 = 21.39 - 0.45 t
t = 47.53 s
(c) Use the formula for the distance traveled in nth second
[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]
where n be the number of second, a be the acceleration, u be the initial velocity.
put n = 1, u = 21.39 m/s , a = - 0.45m/s^2
[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 1-1 \right )}[/tex]
[tex]s_{n^{th}=21.165m[/tex]
(d) Use the formula for the distance traveled in nth second
[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]
where n be the number of second, a be the acceleration, u be the initial velocity.
put n = 5, u = 21.39 m/s , a = - 0.45m/s^2
[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 5-1 \right )}[/tex]
[tex]s_{n^{th}=19.365m[/tex]
