Answer:
The velocity of the car after the collision is -5.36 m/s
Step-by-step explanation:
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.
let the following:
m₁ = mass of the car = 1400 kg
m₂ = mass of the truck = 3200 kg
u₁ = velocity of the car before collision = 13.7 m/s
u₂ = velocity of the truck before collision = 0 m/s
v₁ = velocity of the car after collision
v₂ = velocity of the truck after collision
v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)
= [ 13.7 * (1400 - 3200) + 0 * 2 * 3200 ]/ (1400 + 3200)
= - 5.36 m/s
So, <u>the velocity of the car after the collision is -5.36 m/s</u>
If you have learned how to find the line of best fit manually, then you can do it that way. Perhaps you may want to just find a line that can connect at least two of the points and I believe that that line will be able to represent the other points because, in general, the points are pretty close to one another.
If you don't want to do it manually and have a graphing calculator (which I recommend) then you can use that to find the line of best fit (and if you want then you can see how precise your points are with your r^2 value). Or there is a website (http://illuminations.nctm.org/Activity.aspx?id=4186), which you can use to help you to find the equation of that particular line.
Once you have that done, then you can substitute 2009 for the x value in the equation and then see what y value the equation produces. That will then be your answer :)
Answer:X= 1.84. Y= 3.08
Step-by-step explanation:
-9x+8y =8 ............i
-8x+10y= -16........ii
Multiply equation I by 8
Multiply equation ii by 9
-72x+64y = 64 ..... iii
-72x+90y = 144 ..... iv
Subtract iv from iii
-26y = -80
y= 80/26
y= 3.08
Put the value of y in equation i
-9x+8y =8
-9x+8(3.08) = 8
-9x+24.64 = 8
-9x = 8-24.64
-9x=-16.64
Divide both side by 9
-9x/-9 = -16.64/-9
X= 1.84
Answer:
I think it could be modelled by the function I'm not 100 percent sure but i hope it helps
Step-by-step explanation:
