A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 7.4 s, coasts for 1.8 s, and then slows down at a rate of 3.5 m/s^2 for the next stop sign.How far apart are the stop signs?
To find the distance between the two stop signs, we need to calculate the total distance traveled by the car during each phase of motion and then add them up.
First, let's calculate the distance during the acceleration phase.
The initial velocity, u, is 0 m/s (since the car starts from rest).
The acceleration, a, is 4.6 m/s^2.
The time, t, is 7.4 s.
We can use the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2).
Substituting the values into the formula, distance = (0 * 7.4) + (0.5 * 4.6 * (7.4^2)), simplifying gives distance = 0 + (0.5 * 4.6 * 54.76), and further simplification gives distance = 0 + 125.972 m.
Therefore, the distance covered during the acceleration phase is 125.972 m.
Next, let's calculate the distance during the coasting phase.
The car is coasting, which means there is no acceleration. Therefore, the velocity remains constant. Let's call this velocity v.
The time, t, during coasting is 1.8 s.
We can use the formula: distance = velocity * time.
Since the velocity during coasting remains constant, we can use the velocity from the previous phase (acceleration phase) as the coasting velocity.
Therefore, distance = 4.6 m/s * 1.8 s = 8.28 m.
So, the distance covered during the coasting phase is 8.28 m.
Finally, let's calculate the distance during the deceleration phase.
The final velocity, v, at the end of coasting is the same as the initial velocity for deceleration.
Therefore, the final velocity, v, is 4.6 m/s.
The acceleration, a, during deceleration is -3.5 m/s^2 (negative because it's slowing down).
We need to find the time, t, during deceleration.
We can use the equation: final velocity = initial velocity + (acceleration * time).
Substituting the values: 4.6 = 0 + (-3.5) * time.
Simplifying gives 4.6 = -3.5 * time.
Solving for time, we find time = 4.6 / (-3.5) ≈ -1.314 s (approximately).
Since time cannot be negative in this context, we take the absolute value of time, which is 1.314 s.
Now, we can use the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2).
Substituting the values, distance = (4.6 * 1.314) + (0.5 * -3.5 * (1.314^2)),
simplifying gives distance = 6.0444 + (-2.28086), and further simplification gives distance = 3.76354 m (approximately).
Therefore, the distance covered during the deceleration phase is 3.76354 m.
To find the total distance between the two stop signs, we add up the distance covered during each phase:
Total distance = distance during acceleration + distance during coasting + distance during deceleration.
Total distance = 125.972 m + 8.28 m + 3.76354 m ≈ 137.01554 m (approximately).
Therefore, the two stop signs are approximately 137.01554 meters apart.