Respuesta :
Answer:
[tex]v^2=v_o^2-2\times a\times (v_o.t)[/tex]
Explanation:
Given:
Initial velocity of the vehicle, [tex]v_o[/tex]
distance between the car and the tree, [tex]x_f[/tex]
time taken to respond to the situation, [tex]t[/tex]
acceleration of the car after braking, [tex]a[/tex]
Using equation of motion:
[tex]v^2=u^2+2a.s[/tex] ..............(1)
where:
[tex]v=[/tex] final velocity of the car when it hits the tree
[tex]u=[/tex] initial velocity of the car when the tree falls
[tex]a=[/tex] acceleration after the brakes are applied
[tex]s=[/tex] distance between the tree and the car after the brakes are applied.
[tex]s=v_o\times t[/tex]
Now for this situation the eq. (1) becomes:
[tex]v^2=v_o^2-2\times a\times (v_o.t)[/tex] (negative sign is for the deceleration after the brake is applied to the car.)
