A 1400kg sedan goes through a wide intersection traveling from north to south wh

Question

A 1400kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.39m west and 6.22m south of the impact point.

A.How fast was sedan traveling just before the collision?

B.How fast was SUV traveling just before the collision?

Explanation / Answer

The acceleration after impact is found from
uR = ug(1400 + 2000) =3400a
a = ug = 0.75 x 9.8 = 7.35
The distance slid is
SQRT(5.39^2 + 6.22^2) = 8.230 m
Speed after impact from
v^2 - u^2 = 2as
v = 0
-u^2 = -2 x 7.35 x 8.230
u = 10.99 m/s and angle = tan^-1(6.22/5.39) = 49.08 deg south of west.
Resolving into south and west components and equating momentum before and after
1400v[1] = 3400 x 10.99 x sin(49.08)
2000v[2] = 3400x 10.99 x cos(49.08)
v[1] = 20.167 m/s
v[2] = 12.23 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.