Answer:
[tex]a = -0.54\ m/s^2[/tex]
Explanation:
Given
[tex]u = 23m/s[/tex] --- initial velocity
[tex]v = 0m/s[/tex] --- final velocity
[tex]s = 500m[/tex] --- distance of the car
Required
Determine the train's acceleration to stop 10m before the obstacle
This will be solved using:
[tex]v^2 = u^2 + 2aH[/tex]
Where
[tex]H = s - 10[/tex] i.e to stop 10m before the car
So, we have:
[tex]0^2= 23^2 + 2a(s - 10)[/tex]
[tex]0^2= 23^2 + 2a(500 - 10)[/tex]
[tex]0= 529 + 2a*490[/tex]
Collect like terms
[tex]2a*490 = 0 -529[/tex]
[tex]2a*490 = -529[/tex]
Make a the subject
[tex]a = -\frac{529}{2*490}[/tex]
[tex]a = -\frac{529}{980}[/tex]
[tex]a = -0.54\ m/s^2[/tex] --- approximated
It is negative because the train decelerates
