A car leaves skid marks 90 long on the highway in coming to a stop.
Assuming a deceleration of 3.5m/s^2, estimate the speed of the car just before braking.
0 = Vi - a t
so
t = Vi/3.5
d = Vi t - (1/2)(3.5) t^2
90 = Vi^2/3.5 - 1.75 Vi^2/12.25
Vi^2 - .5 Vi - 315 = 0
Vi = [ .5 +/- sqrt ( .25 + 1260) ]/2
Vi = [ .5 +/- 35.5 ] / 2
Vi = 18 m/s
15
To estimate the speed of the car just before braking, we can use the equation of motion known as the SUVAT equation, which relates the initial velocity (u), final velocity (v), acceleration (a), and distance (s) traveled.
The equation that we will use is:
v^2 = u^2 + 2as
In this case, we know the following information:
The distance traveled (s) = 90 m
The acceleration (a) = -3.5 m/s^2 (negative sign because the car is decelerating)
We need to find the initial velocity (u), which is the speed of the car just before braking. The final velocity (v) is assumed to be zero since the car comes to a stop.
Now, let's rearrange the equation to solve for u:
u^2 = v^2 - 2as
Since v = 0, the equation simplifies to:
u^2 = -2as
Plugging in the values:
u^2 = -2 * (-3.5) * 90
Simplifying further:
u^2 = 630
Taking the square root of both sides:
u = √630
Calculating:
u ≈ 25.1194 m/s
Therefore, the estimated speed of the car just before braking is approximately 25.12 m/s.