Please help me with this question
1 answer:
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Answer:
a = 6.1 m / s²
Explanation:
For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle
Let's find the relative initial velocity of the two vehicles
v₀ = v₀₂ - v₀₁
v₀ = 25.4 - 13.6
v₀ = 11.8 m / s
the fastest vehicle
x = v₀ t + ½ a t²
The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is
x = 11.4 m
let's use the expression
v² = v₀² - 2 a x
how the vehicle stops v = 0
a = v₀² / 2x
a =
a = 6.1 m / s²
this velocity is directed to the left
Answer:
534 ft
Explanation:
given,
speed of the vehicle = 60 mph
1 mph = 0.447 m/s
60 mph = 60 x 0.447 = 26.82 m/s
stopping distance = ?
Stopping distance of the car is equal to the distance traveled in the reaction time and the braking distance.
Reaction time of a common person = 1.5 s
taking coefficient of friction of the road = 0.3
using equation of stopping sight distance
S.D = RD + BD
RD is Reaction distance
RD = v t_r
RD = 26.82 x 1.5
RD = 40.23 m
BD is the braking distance
BD = 122.33 m
Stopping sight distance
SD = 40.23 + 122.33
SD = 162.56 m
1 m = 3.28 ft
162.56 m = 162.56 x 3.28
= 533.20 ft ≈ 534 ft
hence, the stopping distance will be equal to 534 ft
Answer:
Explanation:
The cabinet does not move: this means that the net force acting on it is zero.
Along the horizontal direction, we have two forces:
- The push exerted by Bob, F = 200 N, forward
- The frictional force, , which acts in the opposite direction (backward)
Since the net force must be zero, we have:
So solving the equation we can find the magnitude of the friction force: