Answer:
The stop signs are 216 m apart.
Explanation:
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The equation of position and velocity of the car are the following:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the car at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity of the car at time t.
Let´s place the origin of the frame of reference at the first stop sign so that x0 = 0. Let's calculate the traveled distance during the 6.0 s of acceleration at 4.0 m/s².
x = x0 + v0 · t + 1/2 · a · t² (x0 and v0 = 0)
x = 1/2 · a · t²
x = 1/2 · 4.0 m/s² · (6.0 s)²
x = 72 m
During the first 6.0 s, the car traveled 72 m.
Let's find the velocity reached by the car in that time:
v = v0 + a · t
v = a · t
v = 4.0 m/s · 6.0 m/s²
v = 24 m/s
After the first 6.0 s, the car travels for 2.0 s at v = 24 m/s and a = 0 m/s². The initial position in this case is the distance traveled so far, 72 m. So, after this 2.0 s the position of the car will be:
x = x0 + v · t
x = 72 m + 24 m/s · 2.0 s
x = 120 m
Then the car slows down until it stops at the next stop sign. Let's calculate the time it takes the car to stop, i.e. to reach a velocity of zero:
v = v0 + a · t
When the car stops, the velocity is zero:
0 = 24 m/s - 3.0 m/s² · t (notice that the acceleration is negative because the car is slowing down)
-24 m/s / -3.0 m/s² = t
t = 8.0 s
Then, the car travels 8.0 s slowing down. Let's calculate the position of the car after that time:
x = x0 + v0 · t + 1/2 · a · t²
The initial position and velocity, in this case, are the position and velocity of the car after the previous two parts of the travel:
x = 120 m + 24 m/s · 8.0 s - 1/2 · 3.0 m/s² · (8.0 s)²
x = 216 m
The stop signs are 216 m apart.
The two stop signs are : 216 m apart
Given data :
Acceleration = 4 m/s²
Time ( t₁ ) = 6 secs
Coast time ( t₂ ) = 2 secs
Deceleration = 3 m/s²
Initial conditions :
u₁ = 0 m/s, a₁ = 4 m/s², t₁ = 6 secs,
S₁ = u₁t₁ + 1/2 a₁t₁²
= 0 + 1/2 * 4 * 6²
= 72m
Next step : Determine
S₂ = u₂t₂ + 1/2 a₂t₂² ------ ( 2 )
where: u₂ = v₁ = u₁ + a₁t₁ , a₂ = 0 m/s²
= 0 + 4 * 6 = 24 m/s
back to equation ( 2 )
S₂ = 24 * 2 + 0 = 48 m
Next step : determine
S₃ = u₂t₃ + 1/2a₃t₃² ---- ( 3 )
where : u₂ = 24 m/s , t₃ = 8, a₃ = -3 m/s²,
insert values into equation ( 3 )
S₃ = 24 * 8 + ( 1/2 * (-3 ) * 8²
= 96m
Therefore the two stop signs are = S₁ + S₂ + S₃
= 72 + 48 + 96 = 216 m
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