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
Answer:
I believe the answer is esophagus
Answer:
d A ball is rolling down an inclined plane.
Explanation:
When path length is equal to the displacement
then we can say that the motion of the object must be in straight line so that the distance and displacement must be same
SO here we can say
a A ball on the end of a string is moving in a vertical circle.
In circular path distance and displacement is not same
b A toy train is traveling around a circular track.
In circular path distance and displacement is not same
c A train travels 5 miles east before it stops. It then travels 2 miles west.
Net displacement is 3 miles East while distance is 7 miles
d A ball is rolling down an inclined plane.
Here its motion is in straight line so we can say that path length and displacement will be same
e A ball rises and falls after being thrown straight up from the earth's surface.
In this type of to and fro motion path length is not same as displacement
