Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
A. is the right answer since work is negative and Q which is heat in negative also
Sorry bro I just need points for my calculus exam
Friction produces heat hope this helps
Answer:
a) 2.43 m/s
b) 4.83 m/s
c) 0.023 m/s²
Explanation:
a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration
Speed = Distance / Time

Velocity of car A is 2.43 m/s
t = Time taken = 210 seconds
u = Initial velocity
v = Final velocity
s = Displacement = 510 m
a = Acceleration
c)

Acceleration of car B is 0.023 m/s²
b)

Final velocity of car B is 4.83 m/s