Answer:
a) [tex]t=5\ s[/tex]
b) [tex]s=61.5\ m[/tex]
Explanation:
Given:
acceleration of the car, [tex]a=-4.92\ m.s^{-1}[/tex]
initial velocity of the car, [tex]u=24.6\ m.s^{-1}[/tex]
final velocity of the car, [tex]v=0\ m.s^{-1}[/tex]
a)
Using eq. of motion:
[tex]v=u+a.t[/tex]
[tex]0=24.6-4.92\times t[/tex]
[tex]t=5\ s[/tex]
b)
Distance travelled before stopping:
[tex]s=u.t+\frac{1}{2} a.t^2[/tex]
[tex]s=24.6\times 5-0.5\times 4.92\times 5^2[/tex]
[tex]s=61.5\ m[/tex]
c)
