Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
Answer:
Explanation:
Let the radius of track required be r.
Centripetal force will be provided by frictional force which will be equal to
m v²/ r
Frictional force = mg x μ
So
m v² /r = mg μ
r = v² / μ g =
v = 29 km /h = 8.05 m /s
r =( 8.05 x 8.05 ) /( .32 x 9.8 ) = 20.66 m
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Answer:
I was also going to ask same question edited:ok i found its true
Apply Newton's second law to the car's motion:
F = ma
F = net force, m = mass, a = acceleration
Given values:
F = 2500N, a = 3.0m/s²
Plug in and solve for m:
2500 = m(3.0)
m = 830kg