8/4 = y/y-x
8y - 8x = 4y
y = 2x
y = 2 x 4
y = 8
Hope this helps
Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)
Answer:
Explanation:
See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.
The individual driving times and distances are summed to provide:
(<u>a) How far did he travel?</u>
103 km
<u>(b) What was his average speed?</u>
12.88 km/h
<u>(c) What was his displacement?</u>
73.82 km
<u>(d) What was his average velocity?</u>
9.228 km/h
Answer:
depends on what type of car it is
Explanation: