A car with mass mc = 1225 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck of mass mt = 1654 kg traveling south at vt = 8.6 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of μk = 0.5.
A) Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j.
B) How far, in meters, will the vehicles slide after the collision?

a) v(f) = -4i - 5j

b) 4.18 m

Explanation:

The equation to be used for this question is

v(c)m(c) + v(t)m(t) = [m(c) + m(t)] v(f)

if we rearrange and make v(f) subject of formula, then

v(f) = v(c)m(c) + v(t)m(t) / [m(c) + m(t)]

One vehicle is headed towards south and the other vehicle, west when they collide they will travel together in a southwestern direction. This means that both vehicles are traveling in the negative direction taking a standard frame of reference. Thus, we can write the equation in component form by substituting the values as

v(f) = 1225(-9.5i) + 1654(-8.6j) / 1225 + 1654

v(f) = -11637.5i - 14224.4j / 2879

v(f) = -4i - 5j m/s

From the answer,

v(f) = √(4² + 5²)

v(f) = √41

v(f) = 6.4 m/s

And we know that

KE = ½mv²

Fd = umgd

And, KE = Fd, so

½mv² = umgd

½v² = ugd

Making d the subject of formula,

d = v²/2ug

d = 6.4² / 2 * 0.5 * 9.8

d = 41 / 9.8

d = 4.18 m

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

the answer is a 3.8cm

explanation: