Respuesta :
Answer:
let speed of volvo when slamming brake = v, Mass of volvo = m1
speed of volvo just before coliision = v1
After breaking , skidding distance = 30 m
v1^2- v^2 = - 2 X a X s , a= deceleration and s= skidding distance
Let us assume that the dceleration = -0.3 g = - 2.94 m/s^2 ( Taken from Volvo web-site - for car's emergency system)
v^2 = v1^2 + (2 X a X s )= v1^2 + 176.4
v^2 = v1^2 + 176.4
After collision both cars coalesc and travel together ( m1+m2) with intial speed of V ( say)
Then the combined cars travel 12.25 mrs and come to halt
(0)^2 -(V)^2 = - 2 X a' X s' , where a' = deceeration of both cars combined , s' = distance travelled by them
Let us assume the deceleration of combined system is higher at (- g )
V^2 = 2 X 9.8 X 12.25
V = 15.5 m/s
X- direction ( East) m1 X v1 = (m1+ m2) X V X cos 15 ------ conservation of momentum
v1 = (1650+ 975) Kg X 15.5 m/s X cos 15 /1650 Kg = 23.8 m/s
So v^2= V1^2 + 176.4 = (23.8)^2 + 176.4
v = 27.3 m/s = 61.1 mph ( He is driving beyond speed limit while hitting the brakes )
m2 X v2 = (m1+m2) V sin 15 ------ conservation of momentum for Y-direction
v2 = (m1+m2) X V X sin 15/ m2
v2 = 41.7 m/s ( 93.3 m/hr ) - exceeding speed limit
Answer:
1. the speed of the Volvo is 27.3 m/s
2. the speed of the Alpha Romeo is 10.8 m/s
Explanation:
1. Given:
v = speed of the Volvo
m₁ = mass of the Volvo
v₁ = speed of the Volvo before collision
skidding distance = 30 m
[tex]v_{1} ^{2} -v^{2} =-2a*s[/tex]
where a is the deceleration = -2.94 m/s²
Clearing v
[tex]v^{2} =v_{1} ^{2} +176.4[/tex]
After the collision both cars travel together, then
[tex]0-V^{2} =-2*a*s\\V=\sqrt{2as} =\sqrt{2*9.8*12.25} =12.25m/s[/tex]
The conservation of momentum in x direction
[tex]v_{1} =\frac{(m_{1}+m_{2})*V*cos15 }{m_{1} } =\frac{(1650+975)*15.5*cos15}{1650} =23.8m/s[/tex]
[tex]v=\sqrt{v_{1}^{2}+176.4 } =\sqrt{23.8^{2}+176.4 } =27.3m/s[/tex]
2. The conservation of momentum in y direction
[tex]v_{2} =\frac{(m_{1}+m_{2})*V*sin15 }{m_{2} } =\frac{(1650+975)*15.5*sin15}{975} =10.8m/s[/tex]