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please help 

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 
or 
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!

Guest Oct 5, 2018
 #1
avatar+14445 
+2

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= ma

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)

ElectricPavlov  Oct 5, 2018
edited by ElectricPavlov  Oct 5, 2018

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