Respuesta :
Answer:
The magnitude of the acceleration of the car after the brakes failed is 4 m/s²
Explanation:
The car was originally traveling at 30.0 m/s, that is
The initial velocity, [tex]u[/tex] = 30.0 m/s
The time spent while the car manages to brake is 5.0 seconds, that is
time, [tex]t[/tex] = 5.0 secs
and the distance traveled during this time is
distance, [tex]s[/tex] = 125 m
From one of the equations of kinematics for linear motion,
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]
Where [tex]a[/tex] is the acceleration
We can determine the deceleration of the car during the first 5.0 seconds
Hence,
From,
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]
[tex]125 = 30.0(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]
[tex]125 =150.0 + 12.5a[/tex]
[tex]12.5a = 125 - 150.0[/tex]
[tex]12.5a = -25\\a = \frac{-25}{12.5}\[/tex]
[tex]a = - 2.0 m/s^{2}[/tex] (Negative sign indicates deceleration)
Now we will calculate the final velocity reached at this time
From,
[tex]v^{2} = u^{2} + 2as[/tex]
Where [tex]v[/tex] is the final velocity
[tex]v^{2} = 30.0^{2} + 2(-2.0)(125)\\v^{2} = 400\\v = \sqrt{400} \\v = 20 m/s \\[/tex]
This is the final velocity reached by the car during the first 5.0 seconds
Now, for the magnitude of the acceleration of the car after the brakes failed,
After the brakes failed,
it travels an additional 150 m further down the road, that is
s = 150m
an additional 5.0 seconds, that is
t = 5.0 seconds
Also, from
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]
The initial velocity here will be the final velocity for the first 5.0 seconds, that is,
u = 20 m/s
Hence,
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex] becomes
[tex]150 = 20(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]
[tex]150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = \frac{50}{12.5} \\a = 4m/s^{2}[/tex]
Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²
