A 750 kg van comes to rest from a speed of 75 km/h in a distance of 135 m. Presume the van is initially moving in the positive x direction.
A. If the brakes are the only thing making the van come to a stop, calculate the force(in newtons, in a component along the direction of motion of the car) that ththe brakes apply on the van.
B. Suppose instead of braking, the van hits a concrete abutment at full speed, and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the van in this case.
C. What is he ratio of the force on the van from the concrete to the braking force.
i figured out A. by using F=ma
from newtons third equation
2aS = vf2 - vi2
a = (vf2 - vi2)/2S = -1.6m/s2
F= ma = -1200N
but im too burn out and keep getting the other two wrong. thank you in advance!
a = (vf-vo)/t
re-arranging t = (vf-vo)/a = (0-20.8333)/a = (-20.8333)/a 75 km / hr = 20.8333 m/s
xf = xo + vo t + 1/2 a t^2 Substitute the value of t above
135 = 0 + (20.8333)(-20.8333/a) + 1/2 a (-20.8333/a)^2
135 = -434/a + 1/2 a (434/a^2)
135 = -434/a + 217/a
135 = -217/a a = -1.607 m/s^2
F = 750(-1.607) = -1205 N
B. I find a = -108.49 m/s^2 Use this in F=ma F = -81370 N
C. ratio 81370/1205 = 67.5 times higher
(NOTE : this is the same as the ratio of the stopping distances 135/2 = 67.5)