A car travelling 10m/s increases its speed to 20m/s over a time of 4s. What is the acceleration of the car? A car is traveling a
1 answer:
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Answer:
magnitude of A − B = 15.81 km
Explanation:
Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.
According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.
A(0,-5)
B(15,0)
A - B = (-15 i - 5 j )
Magnitude of the vector is given by
|A - B| =
|A - B| =
|A - B| = 15.81 km
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