Answer:
Decelaration is [tex]13.94m/s^2[/tex]
Time is 4.3 seconds
Explanation:
Given
[tex]u = 60km/hr[/tex] --- Initial Velocity
[tex]s = 10m[/tex] --- Distance to stop
[tex]v = 0km/hr[/tex] -- Final Velocity
Required
Determine the deceleration (a) and time taken (t)
The deceleration is solved using the following Newton equation of motion
[tex]v^2 = u^2 + 2as[/tex]
Convert velocity to m/s
[tex]u = 60km/hr[/tex]
[tex]u = \frac{50 * 1000}{3600} m/s[/tex]
[tex]u = 16.7\ m/s[/tex]
[tex]v^2 = u^2 + 2as[/tex] becomes
[tex]0^2 = 16.7^2 + 2 * a * 10[/tex]
[tex]0 = 278.89+ 20 a[/tex]
Collect Like Terms
[tex]-20a = 278.89[/tex]
Make a the subject
[tex]a = -\frac{278.89}{20}[/tex]
[tex]a = -13.94m/s^2[/tex]
The negative sign shows deceleration.
Hence, the decelaration is [tex]13.94m/s^2[/tex]
Solving time (t): Using first law of motion
[tex]v = u + at[/tex]
Substitute values for v, u and a
[tex]0 = 60 -13.94 * t[/tex]
[tex]0 = 60 -13.94t[/tex]
Collect Like Terms
[tex]13.94t = 60[/tex]
Solve for t
[tex]t = \frac{60}{13.94}[/tex]
[tex]t = 4.3s[/tex]
